Solved Examples: Quadratic Equations

$ x^2 - 6x - 40 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -6 \\[3ex] c = -40 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-6)^2 - 4(1)(-40) \\[3ex] = 36 + 160 \\[3ex] = 196 \\[3ex] $ $196$ is a perfect square.
The quadratic equation can be solved by the Factoring method.
The quadratic equation will have two rational roots.

$ x^2 - 6x - 40 = 0 \\[3ex] (x^2)(-40) = -40x^2 \\[3ex] Factors\: are\: 4x, -10x \\[3ex] x^2 + 4x - 10x - 40 = 0 \\[3ex] x^2 + 4x = x(x + 4) \\[3ex] -10x - 40 = -10(x + 4) \\[3ex] \Rightarrow (x + 4)(x - 10) = 0 \\[3ex] x + 4 = 0 \\[3ex] x = -4 \\[3ex] OR \\[3ex] x - 10 = 0 \\[3ex] x = 10 \\[3ex] \Rightarrow x = -4 \:\:\:OR\:\:\: x = 10 \\[3ex] $ Check

LHS $ x^2 - 6x - 40 \\[3ex] x = -4 \\[3ex] (-4)^2 - 6(-4) - 40 \\[3ex] 16 + 24 - 40 \\[3ex] 40 - 40 \\[3ex] 0 \\[3ex] $ $ x^2 - 6x - 40 \\[3ex] x = 10 \\[3ex] (10)^2 - 6(10) - 40 \\[3ex] 100 - 60 - 40 \\[3ex] 40 - 40 \\[3ex] 0 $

RHS $ 0 $

$ x^2 - 6x - 40 = 0 \\[3ex] x^2 - 6x = 40 \\[3ex] Coefficient\:\: of\:\: x^2 = 1 \\[3ex] Coefficient\:\: of\:\: x = -6 \\[3ex] Half\:\: of\:\: it = -3 \\[3ex] Square\:\: it = (-3)^2 \\[3ex] x^2 - 6x + (-3)^2 = 40 + (-3)^2 \\[3ex] (x - 3)^2 = 40 + 9 \\[3ex] (x - 3)^2 = 49 \\[3ex] (x - 3) = \pm \sqrt{49} \\[3ex] x - 3 = \pm 7 \\[3ex] x = 3 \pm 7 \\[3ex] x = 3 + 7 \:\:\:OR\:\:\: x = 3 - 7 \\[3ex] x = 10 \:\:\:OR\:\:\: x = -4 $

$ x^2 - 6x - 40 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -6 \\[3ex] c = -40 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-40)}}{2(1)} \\[5ex] x = \dfrac{6 \pm \sqrt{36 - (-160)}}{2} \\[5ex] x = \dfrac{6 \pm \sqrt{36 + 160}}{2} \\[5ex] x = \dfrac{6 \pm \sqrt{196}}{2} \\[5ex] x = \dfrac{6 \pm 14}{2} \\[5ex] x = \dfrac{6 + 14}{2} \:\:\:OR\:\:\: x = \dfrac{6 - 14}{2} \\[5ex] x = \dfrac{20}{2} \:\:\:OR\:\:\: x = \dfrac{-8}{2} \\[5ex] x = 10 \:\:\:OR\:\:\: x = -4 \\[3ex] $ Alternative Check
Sum of roots = $\dfrac{-b}{a}$

Sum of roots = $10 + (-4) = 10 - 4 = 6$

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$ \dfrac{-b}{a} = \dfrac{-(-6)}{1} = 6 \\[3ex] $

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Product of roots = $\dfrac{c}{a}$

Product of roots = $10 * (-4) = -40$

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$ \dfrac{c}{a} = \dfrac{-40}{1} = -40 $

$ -x^2 + 2x = 1 \\[3ex] 0 = 1 + x^2 - 2x \\[3ex] x^2 - 2x + 1 = 0 \\[3ex] OR \\[3ex] -x^2 + 2x = 1 \\[3ex] -x^2 + 2x - 1 = 0 \\[3ex] It\: has\: a\: leading\: negative \\[3ex] Factor\: out\: the\: negative \\[3ex] Factor\: by\: GCF \\[3ex] -1(x^2 - 2x + 1) = 0 \\[3ex] Divide\: both\: sides\: by\: -1 \\[3ex] x^2 - 2x + 1 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -2 \\[3ex] c = 1 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-2)^2 - 4(1)(1) \\[3ex] = 4 - 4 \\[3ex] = 0 \\[3ex] $ $0$ is a perfect square.
The quadratic equation can be solved by the Factoring method.
The quadratic equation will have one repeated rational root.

$ -x^2 + 2x = 1 \\[3ex] 0 = 1 + x^2 - 2x \\[3ex] x^2 - 2x + 1 = 0 \\[3ex] (x^2)(1) = x^2 \\[3ex] Factors\: are\: -x, -x \\[3ex] x^2 - x - x + 1 = 0 \\[3ex] x^2 - x = x(x - 1) \\[3ex] -x + 1 = -1(x - 1) \\[3ex] \Rightarrow (x - 1)(x - 1) = 0 \\[3ex] x - 1 = 0 \\[3ex] x = 1 \\[3ex] OR \\[3ex] x - 1 = 0 \\[3ex] x = 1 \\[3ex] \Rightarrow x = 1 \:(twice) \\[3ex] $ Check

LHS $ -x^2 + 2x \\[3ex] -1 * x^2 + 2x \\[3ex] x = 1 \\[3ex] -1 * (1)^2 + 2(1) \\[3ex] - 1 * 1 + 2 \\[3ex] -1 + 2 \\[3ex] 1 $

RHS $ 1 $

$ -x^2 + 2x = 1 \\[3ex] 0 = 1 + x^2 - 2x \\[3ex] x^2 - 2x + 1 = 0 \\[3ex] x^2 - 2x = -1 \\[3ex] Coefficient\:\: of\:\: x^2 = 1 \\[3ex] Coefficient\:\: of\:\: x = -2 \\[3ex] Half\:\: of\:\: it = -1 \\[3ex] Square\:\: it = (-1)^2 \\[3ex] x^2 - 2x + (-1)^2 = -1 + (-1)^2 \\[3ex] (x - 1)^2 = -1 + 1 \\[3ex] (x - 1)^2 = 0 \\[3ex] (x - 1) = \pm \sqrt{0} \\[3ex] x - 1 = \pm 0 \\[3ex] x = 1 \pm 0 \\[3ex] x = 1 + 0 \:\:\:OR\:\:\: x = 1 - 0 \\[3ex] x = 1 \:\:\:OR\:\:\: x = 1 \\[3ex] x = 1 \:(twice) $

$ -x^2 + 2x = 1 \\[3ex] 0 = 1 + x^2 - 2x \\[3ex] x^2 - 2x + 1 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -2 \\[3ex] c = 1 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(1)}}{2(1)} \\[5ex] x = \dfrac{2 \pm \sqrt{4 - 4}}{2} \\[5ex] x = \dfrac{2 \pm \sqrt{0}}{2} \\[5ex] x = \dfrac{2 \pm 0}{2} \\[5ex] x = \dfrac{2 + 0}{2} \:\:\:OR\:\:\: x = \dfrac{2 - 0}{2} \\[5ex] x = \dfrac{2}{2} \:\:\:OR\:\:\: x = \dfrac{2}{2} \\[5ex] x = 1 \:\:\:OR\:\:\: x = 1 \\[3ex] x = 1 \:(twice) \\[3ex] $ Alternative Check
Sum of roots = $\dfrac{-b}{a}$

Sum of roots = $1 + 1 = 2$

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$ \dfrac{-b}{a} = \dfrac{-(-2)}{1} = 2 \\[3ex] $

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Product of roots = $\dfrac{c}{a}$

Product of roots = $1 * 1 = 1$

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$ \dfrac{c}{a} = \dfrac{1}{1} = 1 $

$ -10 = 17x - 20x^2 \\[3ex] -10 - 17x + 20x^2 = 0 \\[3ex] 20x^2 - 17x - 10 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 20 \\[3ex] b = -17 \\[3ex] c = -10 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-17)^2 - 4(20)(-10) \\[3ex] = 289 + 800 \\[3ex] = 1089 \\[3ex] $ $1089$ is a perfect square.
The quadratic equation can be solved by the Factoring method.
The quadratic equation will have two rational roots.

$ 20x^2 - 17x - 10 = 0 \\[3ex] (20x^2)(-10) = -200x^2 \\[3ex] Factors\: are\: -25x, 8x \\[3ex] 20x^2 - 25x + 8x - 10 = 0 \\[3ex] 20x^2 - 25x = 5x(4x - 5) \\[3ex] 8x - 10 = 2(4x - 5) \\[3ex] \Rightarrow (4x - 5)(5x + 2) = 0 \\[3ex] 4x - 5 = 0 \\[3ex] 4x = 5 \\[3ex] x = \dfrac{5}{4} \\[5ex] OR \\[3ex] 5x + 2 = 0 \\[3ex] 5x = -2 \\[3ex] x = -\dfrac{2}{5} \\[5ex] \Rightarrow x = \dfrac{5}{4} \:\:\:OR\:\:\: x = -\dfrac{2}{5} \\[5ex] $ Check

LHS $ -10 $

RHS $ 17x - 20x^2 \\[3ex] x = \dfrac{5}{4} \\[5ex] 17\left(\dfrac{5}{4}\right) - 20\left(\dfrac{5}{4}\right)^2 \\[5ex] = \dfrac{85}{4} - 20\left(\dfrac{25}{16}\right) \\[5ex] = \dfrac{85}{4} - \dfrac{125}{4} \\[5ex] = \dfrac{85 - 125}{4} \\[5ex] -\dfrac{40}{4} \\[5ex] = -10 \\[3ex] $ $ 17x - 20x^2 \\[3ex] x = -\dfrac{2}{5} \\[5ex] 17\left(-\dfrac{2}{5}\right) - 20\left(-\dfrac{2}{5}\right)^2 \\[5ex] = -\dfrac{34}{5} - 20\left(\dfrac{4}{25}\right) \\[5ex] = -\dfrac{34}{5} - \dfrac{16}{5} \\[5ex] = \dfrac{-34 - 16}{4} \\[5ex] -\dfrac{40}{4} \\[5ex] = -10 $

$ -10 = 17x - 20x^2 \\[3ex] -10 - 17x + 20x^2 = 0 \\[3ex] 20x^2 - 17x - 10 = 0 \\[3ex] 20x^2 - 17x = 10 \\[3ex] $ Coefficient of $x^2 = 20$
Make it to be $1$
Divide all terms by $20$

$ \dfrac{20x^2}{20} - \dfrac{17x}{20} = \dfrac{10}{20} \\[5ex] x^2 - \dfrac{17}{20}x = \dfrac{1}{2} \\[5ex] Coefficient\: of\: x = -\dfrac{17}{20} \\[5ex] Half\: of\: it = \dfrac{1}{2} * -\dfrac{17}{20} = -\dfrac{17}{40} \\[5ex] Square\: it = \left(-\dfrac{17}{40}\right)^2 \\[5ex] x^2 - \dfrac{17}{20}x + \left(-\dfrac{17}{40}\right)^2 = \dfrac{1}{2} + \left(-\dfrac{17}{40}\right)^2 \\[5ex] \left(x - \dfrac{17}{40}\right)^2 = \dfrac{1}{2} + \dfrac{289}{1600} \\[5ex] \left(x - \dfrac{17}{40}\right)^2 = \dfrac{800}{1600} + \dfrac{289}{1600} \\[5ex] \left(x - \dfrac{17}{40}\right)^2 = \dfrac{800 + 289}{1600} \\[5ex] \left(x - \dfrac{17}{40}\right)^2 = \dfrac{1089}{1600} \\[5ex] x - \dfrac{17}{40} = \pm \sqrt{\dfrac{1089}{1600}} \\[5ex] x - \dfrac{17}{40} = \pm \dfrac{33}{40} \\[5ex] x = \dfrac{17}{40} \pm \dfrac{33}{40} \\[5ex] x = \dfrac{17 + 33}{40} \:\:\:OR\:\:\: x = \dfrac{17 - 33}{40} \\[5ex] x = \dfrac{50}{40} \:\:\:OR\:\:\: x = -\dfrac{16}{40} \\[5ex] x = \dfrac{5}{4} \:\:\:OR\:\:\: x = -\dfrac{2}{5} $

$ -10 = 17x - 20x^2 \\[3ex] -10 - 17x + 20x^2 = 0 \\[3ex] 20x^2 - 17x - 10 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 20 \\[3ex] b = -17 \\[3ex] c = -10 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-17) \pm \sqrt{(-17)^2 - 4(20)(-10)}}{2(20)} \\[5ex] x = \dfrac{17 \pm \sqrt{289 - (-800)}}{40} \\[5ex] x = \dfrac{17 \pm \sqrt{289 + 800}}{40} \\[5ex] x = \dfrac{17 \pm \sqrt{1089}}{40} \\[5ex] x = \dfrac{17 \pm 33}{40} \\[5ex] x = \dfrac{17 + 33}{40} \:\:\:OR\:\:\: x = \dfrac{17 - 33}{40} \\[5ex] x = \dfrac{50}{40} \:\:\:OR\:\:\: x = -\dfrac{16}{40} \\[5ex] x = \dfrac{5}{4} \:\:\:OR\:\:\: x = -\dfrac{2}{5} \\[5ex] $ Alternative Check
$ -10 = 17x - 20x^2 \\[3ex] -10 - 17x + 20x^2 = 0 \\[3ex] 20x^2 - 17x - 10 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 20 \\[3ex] b = -17 \\[3ex] c = -10 \\[3ex] $ Sum of roots = $\dfrac{-b}{a}$

$ Sum\:\: of\:\: roots \\[3ex] = \dfrac{5}{4} + -\dfrac{2}{5} \\[5ex] = \dfrac{5}{4} - \dfrac{2}{5} \\[5ex] = \dfrac{25}{20} - \dfrac{8}{20} \\[5ex] = \dfrac{25 - 8}{20} \\[5ex] = \dfrac{17}{20} \\[5ex] $

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$ \dfrac{-b}{a} = \dfrac{-(-17)}{20} \\[5ex] = \dfrac{17}{20} \\[5ex] $

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Product of roots = $\dfrac{c}{a}$

$ Product\:\: of\:\: roots \\[3ex] = \dfrac{5}{4} * -\dfrac{2}{5} \\[5ex] = -\dfrac{2}{4} \\[5ex] = -\dfrac{1}{2} \\[5ex] $

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$ \dfrac{c}{a} = \dfrac{-10}{20} \\[5ex] = -\dfrac{1}{2} $

$ -px + 35x^2 = -3 \\[3ex] 35x^2 - px = -3 \\[3ex] 35x^2 - px + 3 = 0 \\[3ex] (35x^2)(3) = 105x^2 \\[3ex] Factors\:\: of\:\: 105x^2 \:\:and\:\: the\:\: sum \\[3ex] 1x \:\:and\:\: 105x = 106x \\[3ex] -1x \:\:and\:\: -105x = -106x \\[3ex] 3x \:\:and\:\: 35x = 38x \\[3ex] -3x \:\:and\:\: -35x = -38x \\[3ex] 5x \:\:and\:\: 21x = 26x \\[3ex] -5x \:\:and\:\: -21x = -26x \\[3ex] 7x \:\:and\:\: 15x = 22x \\[3ex] -7x \:\:and\:\: -15x = -22x \\[3ex] $ So, we can have these equations that can be factored:

$ 35x^2 + 106x + 3 = 0 \\[3ex] 35x^2 - 106x + 3 = 0 \\[3ex] 35x^2 + 38x + 3 = 0 \\[3ex] 35x^2 - 38x + 3 = 0 \\[3ex] 35x^2 + 26x + 3 = 0 \\[3ex] 35x^2 - 26x + 3 = 0 \\[3ex] 35x^2 + 22x + 3 = 0 \\[3ex] 35x^2 - 22x + 3 = 0 \\[3ex] $ The possible values of $p$ are:

$ -p = 106 \implies p = -106 \\[3ex] -p = -106 \implies p = 106 \\[3ex] -p = 38 \implies p = -38 \\[3ex] -p = -38 \implies p = 38 \\[3ex] -p = 26 \implies p = -26 \\[3ex] -p = -26 \implies p = 26 \\[3ex] -p = 22 \implies p = -22 \\[3ex] -p = -22 \implies p = 22 \\[3ex] $

$ (a.) \\[3ex] 2m^2 + 2m - 12 = 0 \\[3ex] 2(m^2 + m - 6) = 0 \\[3ex] m^2 + m - 6 = 0 \\[3ex] Compare:\;\;am^2 + bm + c = 0 \\[3ex] a = 1 \\[3ex] b = 1 \\[3ex] c = -6 \\[3ex] m = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} \\[5ex] = \dfrac{-1 \pm \sqrt{1 + 24}}{2} \\[5ex] = \dfrac{-1 \pm \sqrt{25}}{2} \\[5ex] = \dfrac{-1 \pm 5}{2} \\[5ex] = \dfrac{-1 + 5}{2} \;\;\;OR\;\;\; \dfrac{-1 - 5}{2} \\[5ex] = \dfrac{4}{2} \;\;\;OR\;\;\; \dfrac{-6}{2} \\[5ex] = 2 \;\;\;OR\;\;\; -3 \\[3ex] $ Check

$m = 2$, $m = -3$

LHS	RHS
$ 2m^2 + 2m - 12 \\[3ex] m = 2 \\[3ex] 2(2)^2 + 2(2) - 12 \\[3ex] 8 + 4 - 12 \\[3ex] 0 $ $ m = -3 \\[3ex] 2(-3)^2 + 2(-3) - 12 \\[3ex] 18 - 6 - 12 \\[3ex] 0 $	$0$

$ (b.) \\[3ex] 2x^2 - 3x - 5 = 0 \\[3ex] Compare:\;\;ax^2 + bx + c = 0 \\[3ex] a = 2 \\[3ex] b = -3 \\[3ex] c = -5 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} \\[5ex] = \dfrac{3 \pm \sqrt{9 + 40}}{4} \\[5ex] = \dfrac{3 \pm \sqrt{49}}{4} \\[5ex] = \dfrac{3 \pm 7}{4} \\[5ex] = \dfrac{3 + 7}{4} \;\;\;OR\;\;\; \dfrac{3 - 7}{4} \\[5ex] = \dfrac{10}{4} \;\;\;OR\;\;\; \dfrac{-4}{4} \\[5ex] = \dfrac{5}{2} \;\;\;OR\;\;\; -1 \\[5ex] $ Check

$x = \dfrac{5}{2}$, $x = -1$

LHS	RHS
$ 2x^2 - 3x - 5 \\[3ex] x = \dfrac{5}{2} \\[5ex] 2\left(\dfrac{5}{2}\right)^2 - 3\left(\dfrac{5}{2}\right) - 5 \\[5ex] 2\left(\dfrac{5}{2}\right)\left(\dfrac{5}{2}\right) - \dfrac{15}{2} - 5 \\[5ex] \dfrac{25}{2} - \dfrac{15}{2} - 5 \\[5ex] \dfrac{10}{2} - 5 \\[5ex] 5 - 5 \\[3ex] 0 $ $ x = -1 \\[3ex] 2(-1)^2 - 3(-1) - 5 \\[3ex] 2 + 3 - 5 \\[3ex] 0 $	$0$

$ (c.) \\[3ex] 2x^2 + 3x - 20 = 0 \\[3ex] Compare:\;\;ax^2 + bx + c = 0 \\[3ex] a = 2 \\[3ex] b = 3 \\[3ex] c = -20 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-3 \pm \sqrt{3^2 - 4(2)(-20)}}{2(2)} \\[5ex] = \dfrac{-3 \pm \sqrt{9 + 160}}{4} \\[5ex] = \dfrac{-3 \pm \sqrt{169}}{4} \\[5ex] = \dfrac{-3 \pm 13}{4} \\[5ex] = \dfrac{-3 + 13}{4} \;\;\;OR\;\;\; \dfrac{-3 - 13}{4} \\[5ex] = \dfrac{10}{4} \;\;\;OR\;\;\; \dfrac{-16}{4} \\[5ex] = \dfrac{5}{2} \;\;\;OR\;\;\; -4 \\[5ex] $ Check

$x = \dfrac{5}{2}$, $x = -4$

LHS	RHS
$ 2x^2 + 3x - 20 \\[3ex] x = \dfrac{5}{2} \\[5ex] 2\left(\dfrac{5}{2}\right)^2 + 3\left(\dfrac{5}{2}\right) - 20 \\[5ex] 2\left(\dfrac{5}{2}\right)\left(\dfrac{5}{2}\right) + \dfrac{15}{2} - 20 \\[5ex] \dfrac{25}{2} + \dfrac{15}{2} - 20 \\[5ex] \dfrac{40}{2} - 20 \\[5ex] 20 - 20 \\[3ex] 0 $ $ x = -4 \\[3ex] 2(-4)^2 + 3(-4) - 20 \\[3ex] 32 - 12 - 20 \\[3ex] 0 $	$0$

$ (d.) \\[3ex] 8n^2 - 4n = 18 \\[3ex] 8n^2 - 4n - 18 = 0 \\[3ex] 2(4n^2 - 2n - 9) = 0 \\[3ex] 4n^2 - 2n - 9 = 0 \\[3ex] Compare:\;\;an^2 + bn + c = 0 \\[3ex] a = 4 \\[3ex] b = -2 \\[3ex] c = -9 \\[3ex] n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-9)}}{2(4)} \\[5ex] = \dfrac{2 \pm \sqrt{4 + 144}}{8} \\[5ex] = \dfrac{2 \pm \sqrt{148}}{8} \\[5ex] = \dfrac{2 \pm \sqrt{4 * 37}}{8} \\[5ex] = \dfrac{2 \pm \sqrt{4} * \sqrt{37}}{8} \\[5ex] = \dfrac{2 \pm 2\sqrt{37}}{8} \\[5ex] = \dfrac{2(1 \pm \sqrt{37})}{8} \\[5ex] = \dfrac{1 \pm \sqrt{37}}{4} \\[5ex] = \dfrac{1 + \sqrt{37}}{4} \;\;\;OR\;\;\; \dfrac{1 - \sqrt{37}}{4} \\[5ex] $ Check

$n = \dfrac{1 + \sqrt{37}}{4}$, $n = \dfrac{1 - \sqrt{37}}{4}$

LHS

RHS

$ 8n^2 - 4n \\[3ex] n = \dfrac{1 + \sqrt{37}}{4} \\[5ex] 8\left(\dfrac{1 + \sqrt{37}}{4}\right)^2 - 4\left(\dfrac{1 + \sqrt{37}}{4}\right) \\[5ex] 8 * \dfrac{(1 + \sqrt{37})^2}{4^2} - (1 + \sqrt{37}) \\[5ex] 8 * \dfrac{(1 + \sqrt{37})(1 + \sqrt{37})}{16} - (1 + \sqrt{37}) \\[5ex] \dfrac{1 + \sqrt{37} + \sqrt{37} + 37}{2} - \dfrac{2(1 + \sqrt{37})}{2} \\[5ex] \dfrac{38 + 2\sqrt{37}}{2} - \dfrac{2(1 + \sqrt{37})}{2} \\[5ex] \dfrac{38 + 2\sqrt{37} - 2(1 + \sqrt{37})}{2} \\[5ex] \dfrac{38 + 2\sqrt{37} - 2 - 2\sqrt{37}}{2} \\[5ex] \dfrac{36}{2} \\[5ex] 18 $ $ n = \dfrac{1 - \sqrt{37}}{4} \\[5ex] 8\left(\dfrac{1 - \sqrt{37}}{4}\right)^2 - 4\left(\dfrac{1 - \sqrt{37}}{4}\right) \\[5ex] 8 * \dfrac{(1 - \sqrt{37})^2}{4^2} - (1 - \sqrt{37}) \\[5ex] 8 * \dfrac{(1 - \sqrt{37})(1 - \sqrt{37})}{16} - (1 - \sqrt{37}) \\[5ex] \dfrac{1 - \sqrt{37} - \sqrt{37} + 37}{2} - \dfrac{2(1 - \sqrt{37})}{2} \\[5ex] \dfrac{38 - 2\sqrt{37}}{2} - \dfrac{2(1 - \sqrt{37})}{2} \\[5ex] \dfrac{38 - 2\sqrt{37} - 2(1 - \sqrt{37})}{2} \\[5ex] \dfrac{38 - 2\sqrt{37} - 2 + 2\sqrt{37}}{2} \\[5ex] \dfrac{36}{2} \\[5ex] 18 $

$18$

$ (e.) \\[3ex] 4a^2 - 8 = a \\[3ex] Let\;\;a = p \;\;(to\;\;avoid\;\;conflict\;\;with\;\;the\;\;variable:a\;\;in\;\;the\;\;Quadratic\;\;Formula) \\[3ex] 4p^2 - 8 = p \\[3ex] 4p^2 - p - 8 = 0 \\[3ex] Compare:\;\;ap^2 + bp + c = 0 \\[3ex] a = 4 \\[3ex] b = -1 \\[3ex] c = -8 \\[3ex] p = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-8)}}{2(4)} \\[5ex] = \dfrac{1 \pm \sqrt{1 + 128}}{8} \\[5ex] = \dfrac{1 \pm \sqrt{129}}{8} \\[5ex] = \dfrac{1 + \sqrt{129}}{8} \;\;\;OR\;\;\; \dfrac{1 - \sqrt{129}}{8} \\[5ex] But:\;\; we\;\;let\;\;a = p \\[3ex] Resubstitute:\;\; p = a \\[3ex] \therefore a = \dfrac{1 + \sqrt{129}}{8} \;\;\;OR\;\;\; \dfrac{1 - \sqrt{129}}{8} \\[5ex] $ Check

$a = \dfrac{1 + \sqrt{129}}{8}$, $a = \dfrac{1 - \sqrt{129}}{8}$

LHS

RHS

$ 4a^2 - 8 \\[3ex] a = \dfrac{1 + \sqrt{129}}{8} \\[5ex] 4\left(\dfrac{1 + \sqrt{129}}{8}\right)^2 - 8 \\[5ex] 4 * \left(\dfrac{1 + \sqrt{129}}{8}\right)\left(\dfrac{1 + \sqrt{129}}{8}\right) - 8 \\[5ex] \left(\dfrac{1 + \sqrt{129}}{2}\right)\left(\dfrac{1 + \sqrt{129}}{8}\right) - 8 \\[5ex] \dfrac{(1 + \sqrt{129})(1 + \sqrt{129})}{16} - 8 \\[5ex] \dfrac{1 + \sqrt{129} + \sqrt{129} + 129}{16} - \dfrac{128}{16} \\[5ex] \dfrac{130 + 2\sqrt{129}}{16} - \dfrac{128}{16} \\[5ex] \dfrac{130 + 2\sqrt{129} - 128}{16} \\[5ex] \dfrac{2 + 2\sqrt{129}}{16} \\[5ex] \dfrac{2(1 + \sqrt{129)}}{16} \\[5ex] \dfrac{1 + \sqrt{129}}{8} $ $ a = \dfrac{1 - \sqrt{129}}{8} \\[5ex] 4\left(\dfrac{1 - \sqrt{129}}{8}\right)^2 - 8 \\[5ex] 4 * \left(\dfrac{1 - \sqrt{129}}{8}\right)\left(\dfrac{1 - \sqrt{129}}{8}\right) - 8 \\[5ex] \left(\dfrac{1 - \sqrt{129}}{2}\right)\left(\dfrac{1 - \sqrt{129}}{8}\right) - 8 \\[5ex] \dfrac{(1 - \sqrt{129})(1 - \sqrt{129})}{16} - 8 \\[5ex] \dfrac{1 - \sqrt{129} - \sqrt{129} + 129}{16} - \dfrac{128}{16} \\[5ex] \dfrac{130 - 2\sqrt{129}}{16} - \dfrac{128}{16} \\[5ex] \dfrac{130 - 2\sqrt{129} - 128}{16} \\[5ex] \dfrac{2 - 2\sqrt{129}}{16} \\[5ex] \dfrac{2(1 - \sqrt{129)}}{16} \\[5ex] \dfrac{1 - \sqrt{129}}{8} $

$ a \\[3ex] \dfrac{1 + \sqrt{129}}{8} $ $ \dfrac{1 - \sqrt{129}}{8} $

$ (f.) \\[3ex] 3a^2 - a - 12 = 0 \\[3ex] Let\;\;a = k \;\;(to\;\;avoid\;\;conflict\;\;with\;\;the\;\;variable:a\;\;in\;\;the\;\;Quadratic\;\;Formula) \\[3ex] 3k^2 - k - 12 = 0 \\[3ex] Compare:\;\;an^2 + bn + c = 0 \\[3ex] a = 3 \\[3ex] b = -1 \\[3ex] c = -12 \\[3ex] k = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-12)}}{2(3)} \\[5ex] = \dfrac{1 \pm \sqrt{1 + 144}}{6} \\[5ex] = \dfrac{1 \pm \sqrt{145}}{6} \\[5ex] = \dfrac{1 + \sqrt{145}}{6} \;\;\;OR\;\;\; \dfrac{1 - \sqrt{145}}{6} \\[5ex] But:\;\; we\;\;let\;\;a = k \\[3ex] Resubstitute:\;\; k = a \\[3ex] \therefore a = \dfrac{1 + \sqrt{145}}{6} \;\;\;OR\;\;\; \dfrac{1 - \sqrt{145}}{6} \\[5ex] $ Check

$a = \dfrac{1 + \sqrt{145}}{6}$, $a = \dfrac{1 - \sqrt{145}}{6}$

LHS

RHS

$ 3a^2 - a - 12 \\[3ex] a = \dfrac{1 + \sqrt{145}}{6} \\[5ex] 3\left(\dfrac{1 + \sqrt{145}}{6}\right)^2 - \left(\dfrac{1 + \sqrt{145}}{6}\right) - 12 \\[5ex] 3 * \dfrac{1}{6} * \dfrac{1}{6} * (1 + \sqrt{145})(1 + \sqrt{145}) - \left(\dfrac{1 + \sqrt{145}}{6}\right) - 12 \\[5ex] \dfrac{1}{12} * (1 + \sqrt{145} + \sqrt{145} + 145) - 2\left(\dfrac{1 + \sqrt{145}}{12}\right) - \dfrac{144}{12} \\[5ex] \dfrac{1 + 2\sqrt{145} + 145}{12} - 2\left(\dfrac{1 + \sqrt{145}}{12}\right) - \dfrac{144}{12} \\[5ex] \dfrac{146 + 2\sqrt{145} - 2(1 + \sqrt{145}) - 144}{12} \\[5ex] \dfrac{146 + 2\sqrt{145} - 2 - 2\sqrt{145} - 144}{12} \\[5ex] \dfrac{0}{12} \\[5ex] 0 $ $ a = \dfrac{1 - \sqrt{145}}{6} \\[5ex] 3\left(\dfrac{1 - \sqrt{145}}{6}\right)^2 - \left(\dfrac{1 - \sqrt{145}}{6}\right) - 12 \\[5ex] 3 * \dfrac{1}{6} * \dfrac{1}{6} * (1 - \sqrt{145})(1 - \sqrt{145}) - \left(\dfrac{1 - \sqrt{145}}{6}\right) - 12 \\[5ex] \dfrac{1}{12} * (1 - \sqrt{145} - \sqrt{145} + 145) - 2\left(\dfrac{1 - \sqrt{145}}{12}\right) - \dfrac{144}{12} \\[5ex] \dfrac{1 - 2\sqrt{145} + 145}{12} - 2\left(\dfrac{1 - \sqrt{145}}{12}\right) - \dfrac{144}{12} \\[5ex] \dfrac{146 - 2\sqrt{145} - 2(1 - \sqrt{145}) - 144}{12} \\[5ex] \dfrac{146 - 2\sqrt{145} - 2 + 2\sqrt{145} - 144}{12} \\[5ex] \dfrac{0}{12} \\[5ex] 0 $

$0$

$ f(x) = 3^{x^2 - x - 2} \\[3ex] But\:\: f(x) = 1 \\[3ex] 1 = 3^{x^2 - x - 2} \\[3ex] 3^{x^2 - x - 2} = 1 \\[3ex] 1 = 3^0 ...Law\;3...Exp \\[3ex] \implies \\[3ex] 3^{x^2 - x - 2} = 3^0 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x^2 - x - 2 = 0 \\[3ex] (x + 1)(x - 2) = 0 \\[3ex] x + 1 = 0 \:\:OR\:\: x - 2 = 0 \\[3ex] x = -1 \:\:OR\:\: x = 2 \\[3ex] $

We can do this in two ways.
First Method: Calculate the Discriminant
This method is recommended for the ACT.

$ 5x^2 + 7x + 2 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 5 \\[3ex] b = 7 \\[3ex] c = 2 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = 7^2 - 4(5)(2) \\[3ex] = 49 - 40 \\[3ex] = 4 \\[3ex] $ The discriminant is a perfect square.
The equation has 2 rational roots.

Second Method: Try to Solve it
Check to see if it can be solved by Factoring

$ 5x^2 + 7x + 2 = 0 \\[3ex] (5x^2)(2) = 10x^2 \\[3ex] Factors\:\: are\:\: 5x \:\:and\:\: 2x \\[3ex] 5x^2 + 5x + 2x + 2 = 0 \\[3ex] 5x(x + 1) + 2(x + 1) = 0 \\[3ex] (x + 1)(5x + 2) = 0 \\[3ex] $ It can be solved by Factoring.
The factors are not the same.
The equation has 2 rational roots.

$4x^3 - 2x^2 + x + 7 = 0$
One solution is $x = -1$
To find the other two solutions:
First: Let us use Synthetic Division to find the quotient.
You can learn about Synthetic Division here
That quotient should be quadratic (division of a cubic by a linear)

$ \begin{array}{c|c c c c} -1 & 4 & -2 & 1 & 7 \\[2ex] \hline & & -4 & 6 & -7 \\[2ex] \hline & 4 & -6 & 7 & 0 \end{array} \\[3ex] $ The quotient is $4x^2 - 6x + 7$

So, $4x^2 - 6x + 7 = 0$
Second: Calculate the Discriminant

$ 4x^2 - 6x + 7 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 4 \\[3ex] b = -6 \\[3ex] c = 7 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-6)^2 - 4(4)(7) \\[3ex] = 36 - 112 \\[3ex] = -76 \\[3ex] $ The discriminant is a negative number.
The equation has $2$ complex roots.
Both are complex numbers that are not real.

$ 5x^2 + 6x = -9 \\[3ex] 5x^2 + 6x + 9 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 5 \\[3ex] b = 6 \\[3ex] c = 9 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (6)^2 - 4(5)(9) \\[3ex] = 36 - 180 \\[3ex] = -144 \\[3ex] $ The discriminant is negative.
The quadratic equation cannot be solved by the Factoring method.
The quadratic equation will have two complex roots.
We are going to solve it using the Completing the Square method.

$5x^2 + 6x = -9 \\[3ex]$ Coefficient of $x^2 = 5$
Make it to be $1$
Divide all terms by $5$

$ \dfrac{5x^2}{5} + \dfrac{6x}{5} = -\dfrac{9}{5} \\[5ex] x^2 + \dfrac{6}{5}x = -\dfrac{9}{5} \\[5ex] Coefficient\:\: of\:\: x = \dfrac{6}{5} \\[5ex] Half\:\: of\:\: it = \dfrac{1}{2} * \dfrac{6}{5} = \dfrac{3}{5} \\[5ex] Square\: it = \left(\dfrac{3}{5}\right)^2 \\[5ex] x^2 + \dfrac{6}{5}x + \left(\dfrac{3}{5}\right)^2 = -\dfrac{9}{5} + \left(\dfrac{3}{5}\right)^2 \\[5ex] \left(x + \dfrac{3}{5}\right)^2 = -\dfrac{9}{5} + \dfrac{9}{25} \\[5ex] \left(x + \dfrac{3}{5}\right)^2 = -\dfrac{45}{25} + \dfrac{9}{25} \\[5ex] \left(x + \dfrac{3}{5}\right)^2 = \dfrac{-45 + 9}{25} \\[5ex] \left(x + \dfrac{3}{5}\right)^2 = -\dfrac{36}{25} \\[5ex] x + \dfrac{3}{5} = \pm \sqrt{-\dfrac{36}{25}} \\[5ex] x + \dfrac{3}{5} = \pm \dfrac{6}{5}i \\[5ex] x = -\dfrac{3}{5} \pm \dfrac{6}{5}i \\[5ex] x = \dfrac{-3 \pm 6i}{5} \\[5ex] x = \dfrac{-3 + 6i}{5} \:\:\:OR\:\:\: x = \dfrac{-3 - 6i}{5} \\[5ex] $ Check

LHS $ 5x^2 + 6x \\[3ex] x = \dfrac{-3 + 6i}{5} \\[5ex] x^2 = \left(\dfrac{-3 + 6i}{5}\right)^2 \\[5ex] x^2 = \dfrac{(-3 + 6i)^2}{5^2} \\[5ex] (-3 + 6i)^2 = (-3 + 6i)(-3 + 6i) \\[3ex] (-3 + 6i)^2 = 9 - 18i - 18i + 36i^2 \\[3ex] (-3 + 6i)^2 = 9 - 36i + 36(-1) \\[3ex] (-3 + 6i)^2 = 9 - 36i - 36 \\[3ex] (-3 + 6i)^2 = -27 - 36i \\[3ex] \implies x^2 = \dfrac{-27 - 36i}{25} \\[5ex] 5x^2 = 5\left(\dfrac{-27 - 36i}{25}\right) \\[5ex] 5x^2 = \dfrac{-27 - 36i}{5} \\[5ex] 6x = 6\left(\dfrac{-3 + 6i}{5}\right) \\[5ex] 6x = \dfrac{-18 + 36i}{5} \\[5ex] \implies 5x^2 + 6x \\[3ex] = \left(\dfrac{-27 - 36i}{5}\right) + \left(\dfrac{-18 + 36i}{5}\right) \\[5ex] = \dfrac{(-27 - 36i) + (-18 + 36i)}{5} \\[5ex] = \dfrac{-27 - 36i - 18 + 36i}{5} \\[5ex] = \dfrac{-45}{5} \\[5ex] = -9 $ $ 5x^2 + 6x \\[3ex] x = \dfrac{-3 - 6i}{5} \\[5ex] x^2 = \left(\dfrac{-3 + 6i}{5}\right)^2 \\[5ex] x^2 = \dfrac{(-3 - 6i)^2}{5^2} \\[5ex] (-3 - 6i)^2 = (-3 - 6i)(-3 - 6i) \\[3ex] (-3 - 6i)^2 = 9 + 18i + 18i + 36i^2 \\[3ex] (-3 - 6i)^2 = 9 + 36i + 36(-1) \\[3ex] (-3 - 6i)^2 = 9 + 36i - 36 \\[3ex] (-3 - 6i)^2 = -27 + 36i \\[3ex] \implies x^2 = \dfrac{-27 + 36i}{25} \\[5ex] 5x^2 = 5\left(\dfrac{-27 + 36i}{25}\right) \\[5ex] 5x^2 = \dfrac{-27 + 36i}{5} \\[5ex] 6x = 6\left(\dfrac{-3 - 6i}{5}\right) \\[5ex] 6x = \dfrac{-18 - 36i}{5} \\[5ex] \implies 5x^2 + 6x \\[3ex] = \left(\dfrac{-27 + 36i}{5}\right) + \left(\dfrac{-18 - 36i}{5}\right) \\[5ex] = \dfrac{(-27 + 36i) + (-18 - 36i)}{5} \\[5ex] = \dfrac{-27 + 36i - 18 - 36i}{5} \\[5ex] = \dfrac{-45}{5} \\[5ex] = -9 $

RHS $ -9 $

$ ax^2 + bx + c = 0 \\[3ex] 5x^2 + 6x = -9 \\[3ex] 5x^2 + 6x + 9 = 0 \\[3ex] a = 5 \\[3ex] b = 6 \\[3ex] c = 9 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-6 \pm \sqrt{(6)^2 - 4(5)(9)}}{2(5)} \\[5ex] x = \dfrac{-6 \pm \sqrt{36 - 180}}{10} \\[5ex] x = \dfrac{-6 \pm \sqrt{-144}}{10} \\[5ex] x = \dfrac{-6 \pm \sqrt{-1 * 144}}{10} \\[5ex] x = \dfrac{-6 \pm \sqrt{-1} * \sqrt{144}}{10} \\[5ex] x = \dfrac{-6 \pm i * 12}{10} \\[5ex] x = \dfrac{-6 \pm 12i}{10} \\[5ex] x = \dfrac{2(-3 \pm 6i)}{10} \\[5ex] x = \dfrac{-3 \pm 6i}{5} \\[5ex] x = \dfrac{-3 + 6i}{5} \:\:\:OR\:\:\: x = \dfrac{-3 - 6i}{5} \\[5ex] $ Alternative Check
$ 5x^2 + 6x = -9 \\[3ex] 5x^2 + 6x + 9 = 0 \\[3ex] a = 5 \\[3ex] b = 6 \\[3ex] c = 9 \\[3ex] $ Sum of roots = $\dfrac{-b}{a}$

$ Sum\:\: of\:\: roots \\[3ex] = \dfrac{-3 + 6i}{5} + \dfrac{-3 - 6i}{5} \\[5ex] = \dfrac{(-3 + 6i) + (-3 - 6i)}{5} \\[5ex] = \dfrac{-3 + 6i - 3 - 6i}{5} \\[5ex] = -\dfrac{6}{5} \\[5ex] $

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$\dfrac{-b}{a} = \dfrac{-6}{5} \\[5ex]$

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Product of roots = $\dfrac{c}{a}$

$ Product\:\: of\:\: roots \\[3ex] = \left(\dfrac{-3 + 6i}{5}\right)\left(\dfrac{-3 - 6i}{5}\right) \\[5ex] = \dfrac{(-3 + 6i)(-3 - 6i)}{5^2} \\[5ex] (-3 + 6i)(-3 - 6i) = (-3)^2 - (6i)^2 \\[3ex] (-3 + 6i)(-3 - 6i) = 9 - (6^2 * i^2) \\[3ex] (-3 + 6i)(-3 - 6i) = 9 - (36 * i^2) \\[3ex] (-3 + 6i)(-3 - 6i) = 9 - (36 * -1) \\[3ex] (-3 + 6i)(-3 - 6i) = 9 - (-36) \\[3ex] (-3 + 6i)(-3 - 6i) = 9 + 36 = 45 \\[3ex] = \dfrac{45}{25} \\[5ex] = \dfrac{9}{5} \\[5ex] $

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$\dfrac{c}{a} = \dfrac{9}{5}$

$ (a.) \\[3ex] m^2 - 5m - 14 = 0 \\[3ex] (m + 2)(m - 7) = 0 \\[3ex] m + 2 = 0 \;\;\;OR\;\;\; m - 7 = 0 \\[3ex] m = -2 \;\;\;OR\;\;\; m = 7 \\[3ex] $ Check

$m = -2, \;\;\; m = 7$

LHS	RHS
$ m = -2 \\[3ex] m^2 - 5m - 14 \\[3ex] (-2)^2 - 5(-2) - 14 \\[3ex] 4 + 10 - 14 \\[3ex] 0 $ $ m = 7 \\[3ex] m^2 - 5m - 14 \\[3ex] 7^2 - 5(7) - 14 \\[3ex] 49 - 35 - 14 \\[3ex] 0 $	$0$ $0$

$ (b.) \\[3ex] x^2 + 4x + 3 = 0 \\[3ex] (x + 3)(x + 1) = 0 \\[3ex] x + 3 = 0 \;\;\;OR\;\;\; x + 1 = 0 \\[3ex] x = -3 \;\;\;OR\;\;\; x = -1 \\[3ex] $ Check

$x = -3, \;\;\; x = -1$

LHS	RHS
$ x = -3 \\[3ex] x^2 + 4x + 3 \\[3ex] (-3)^2 + 4(-3) + 3 \\[3ex] 9 - 12 + 3 \\[3ex] 0 $ $ x = -1 \\[3ex] x^2 + 4x + 3 \\[3ex] (-1)^2 + 4(-1) + 3 \\[3ex] 1 - 4 + 3 \\[3ex] 0 $	$0$ $0$

$ (c.) \\[3ex] 2m^2 + 2m - 12 = 0 \\[3ex] 2(m^2 + m - 6) = 0 \\[3ex] m^2 + m - 6 = 0 \\[3ex] (m + 3)(m - 2) = 0 \\[3ex] m + 3 = 0 \;\;\;OR\;\;\; m - 2 = 0 \\[3ex] m = -3 \;\;\;OR\;\;\; m = 2 \\[3ex] $ Check

$m = -3, \;\;\; m = 2$

LHS	RHS
$ m = -3 \\[3ex] 2m^2 + 2m - 12 \\[3ex] 2(-3)^2 + 2(-3) - 12 \\[3ex] 2(9) - 6 - 12 \\[3ex] 18 - 6 - 12 \\[3ex] 0 $ $ m = 2 \\[3ex] 2m^2 + 2m - 12 \\[3ex] 2(2)^2 + 2(2) - 12 \\[3ex] 2^3 + 4 - 12 \\[3ex] 8 + 4 - 12 \\[3ex] 0 $	$0$ $0$

$ \underline{Factoring} \\[3ex] x^2 - x = 12 \\[3ex] x^2 - x - 12 = 0 \\[3ex] (x + 3)(x - 4) = 0 \\[3ex] x + 3 = 0 \;\;OR\;\; x - 4 = 0 \\[3ex] x = -3 \;\;OR\;\; x = 4 \\[3ex] Solution\;\;set = \{-3, 4\} $

For both solutions of the equation to be positive integers, this means that the equation can be solved by the Factoring method.

$ x^2 + kx + 17 = 0 \\[3ex] (x^2)(17) = 17x^2 \\[3ex] Factors\:\: are: \\[3ex] x \:\:and\:\: 17x = 18x...two\:\: negative\:\: solutions \\[3ex] -x \:\:and\:\: -17x = -18x...two\:\: positive\:\: solutions \\[3ex] \implies (x - 1)(x - 17) = 0 \\[3ex] \implies kx = -18x \\[3ex] k = -18 $

$ 2^{x^2 + 1} = 1 \\[3ex] 2^{x^2 + 1} = 2^0 \\[3ex] x^2 + 1 = 0 \\[3ex] x^2 = -1 \\[3ex] x = \pm \sqrt{-1} \\[3ex] $ $2$ imaginary numbers

$ x^2 + 4 = 29 \\[3ex] x^2 = 29 - 4 \\[3ex] x^2 = 25 \\[3ex] x^2 - 4 = 25 - 4 = 21 $

We can solve it in two ways
Use any way that is easier and faster for you

$ \underline{First\:\:Method:\;\; Square\:\:Root\:\:Property} \\[3ex] 4x^2 - 9 = 0 \\[3ex] 4x^2 = 9 \\[3ex] x^2 = \dfrac{9}{4} \\[5ex] x = \pm \sqrt{\dfrac{9}{4}} \\[5ex] x = \pm \dfrac{3}{2} \\[5ex] \underline{Second\:\:Method:\;\; Multiple\:\:Methods} \\[3ex] 4x^2 - 9 = 0 \\[3ex] \underline{Difference\:\:of\:\:Two\:\:Squares} \\[3ex] 2^2x^2 - 3^2 = 0 \\[3ex] (2x)^2 - 3^2 = 0 \\[3ex] (2x + 3)(2x - 3) = 0 \\[3ex] \underline{Zero\:\:Product\:\:Property} \\[3ex] 2x + 3 = 0 \:\:OR\:\: 2x - 3 = 0 \\[3ex] 2x = -3 \:\:OR\:\: 2x = 3 \\[3ex] x = -\dfrac{3}{2} \:\:OR\:\: x = \dfrac{3}{2} $

$ Equal\:\:roots \implies Discriminant = 0 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] x^2 - (p - 2)x + 2p + 1 = 0 \\[3ex] Compare:\:\: ax^2 + bx + c = 0 \\[3ex] \rightarrow a = 1 \\[3ex] b = p - 2 \\[3ex] c = 2p + 1 \\[3ex] b^2 = (p - 2)^2 = (p - 2)(p - 2) \\[3ex] b^2 = p^2 - 2p - 2p + 4 \\[3ex] b^2 = p^2 - 4p + 4 \\[3ex] 4ac = 4(1)(2p + 1) \\[3ex] 4ac = 4(2p + 1) \\[3ex] 4ac = 8p + 4 \\[3ex] b^2 - 4ac = p^2 - 4p + 4 - (8p + 4) \\[3ex] Discriminant = p^2 - 4p + 4 - 8p - 4 \\[3ex] Discriminant = p^2 - 12p \\[3ex] \rightarrow p^2 - 12p = 0 \\[3ex] p(p - 12) = 0 \\[3ex] p = 0 \:\:\:OR\:\:\: p - 12 = 0 \\[3ex] p = 0 \:\:\:OR\:\:\: p = 12 $

$(2, -5)$ implies that $x = 2, y = -5$
The ordered pair is a solution if the LHS = RHS when those values are substituted in the equation.

$ -x^2 - y^2 = -29 \\[3ex] \underline{LHS} \\[3ex] -x^2 - y^2 \\[3ex] Substitute\:\:the\:\:variables\:\:with\:\:those\:\:values \\[3ex] -1 * x^2 - 1 * y^2 \\[3ex] -1 * (2)^2 - 1 * (-5)^2 \\[3ex] -1 * 4 - 1 * 25 \\[3ex] -4 - 25 \\[3ex] -29 \\[3ex] \underline{RHS} \\[3ex] -29 \\[3ex] LHS = RHS \\[3ex] \therefore (2, -5)\:\:is\:\:a\:\:solution. $

$ x^2 = - 12x - 20 \\[3ex] x^2 + 12x + 20 = 0 \\[3ex] \underline{Factoring} \\[3ex] x^2 * 20 = 20x^2 \\[3ex] Factors\;\;are\;\;2x\;\;and\;\;10x \\[3ex] \implies \\[3ex] x^2 + 2x + 10x + 20 = 0 \\[3ex] x(x + 2) + 10(x + 2) = 0 \\[3ex] (x + 2)(x + 10) = 0 \\[3ex] x + 2 = 0 \;\;\;OR\;\;\; x + 10 = 0 \\[3ex] x = -2 \;\;\;OR\;\;\; x = -10 \\[5ex] \underline{Completing\;\;the\;\;Square} \\[3ex] x^2 + 12x + 20 = 0 \\[3ex] x^2 + 12x = -20 \\[3ex] Coefficient\;\;x = 12 \\[3ex] Half\;\;of\;\;it = \dfrac{1}{2} * 12 = 6 \\[5ex] Square\;\;it = 6^2 \\[3ex] Add\;\;to\;\;both\;\;sides \\[3ex] \implies \\[3ex] x^2 + 12x + 6^2 = -20 + 6^2 \\[3ex] (x + 6)^2 = -20 + 36 \\[3ex] (x + 6)^2 = 16 \\[3ex] x + 6 = \pm \sqrt{16} \\[3ex] x + 6 = \pm 4 \\[3ex] x = -6 \pm 4 \\[3ex] x = -6 + 4 \;\;\;OR\;\;\; x = -6 - 4 \\[3ex] x = -2 \;\;\;OR\;\;\; x = -10 \\[5ex] \underline{Quadratic\;\;Formula} \\[3ex] x^2 + 12x + 20 = 0 \\[3ex] a = 1 \\[3ex] b = 12 \\[3ex] c = 20 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-12 \pm \sqrt{12^2 - 4(1)(20)}}{2(1)} \\[5ex] x = \dfrac{-12 \pm \sqrt{144 - 80}}{2} \\[5ex] x = \dfrac{-12 \pm \sqrt{64}}{2} \\[5ex] x = \dfrac{-12 \pm 8}{2} \\[5ex] x = \dfrac{-12 + 8}{2} \;\;\;OR\;\;\; \dfrac{-12 - 8}{2} \\[5ex] x = -\dfrac{4}{2} \;\;\;OR\;\;\; -\dfrac{20}{2} \\[5ex] x = -2 \;\;\;OR\;\;\; x = -10 \\[3ex] $

Check: $x = -2$ OR $x = -10$

LHS	RHS
$x^2$	$- 12x - 20$
$ (-2)^2 \\[3ex] 4 \\[3ex] $ $ (-10)^2 \\[3ex] 100 $	$ -12(-2) - 20 \\[3ex] 24 - 20 \\[3ex] 4 \\[3ex] $ $ -12(-10) - 20 \\[3ex] 120 - 20 \\[3ex] 100 $

$ 3x^2 - 5x = 2 \\[3ex] \underline{1st\;\;Method:\;\;Factoring} \\[3ex] 3x^2 - 5x - 2 = 0 \\[3ex] 3x^2(-2) = -6x^2 \\[3ex] Factors\;\;are:\;\;x\;\;and\;\;-6x \\[3ex] 3x^2 + x - 6x - 2 = 0 \\[3ex] x(3x + 1) -2(3x + 1) = 0 \\[3ex] 3x + 1 = 0 \;\;OR\;\; x - 2 = 0 \\[3ex] 3x + 1 = 0 \\[3ex] 3x = -1 \\[3ex] x = - \dfrac{1}{3} \\[5ex] x - 2 = 0 \\[3ex] x = 2 \\[3ex] x = -\dfrac{1}{3} \;\;OR\;\; x = 2 \\[5ex] \underline{2nd\;\;Method:\;\; Completing\;\;the\;\;Square} \\[3ex] 3x^2 - 5x = 2 \\[3ex] Divide\;\;all\;\;terms\;\;by\;\;3 \\[3ex] \dfrac{3x^2}{3} - \dfrac{5}{3}x = \dfrac{2}{3} \\[5ex] x^2 - \dfrac{5}{3}x = \dfrac{2}{3} \\[5ex] Coefficient\;\;of\;\;x = -\dfrac{5}{3} \\[5ex] Half\;\;of\;\;it = \dfrac{1}{2} * -\dfrac{5}{3} = -\dfrac{5}{6} \\[5ex] Square\;\;it = \left(-\dfrac{5}{6}\right)^2 \\[5ex] Add\;\;to\;\;both\;\;sides \\[3ex] x^2 - \dfrac{5}{3}x + \left(-\dfrac{5}{6}\right)^2 = \dfrac{2}{3} + \left(-\dfrac{5}{6}\right)^2 \\[5ex] \left(x -\dfrac{5}{6}\right)^2 = \dfrac{2}{3} + \dfrac{25}{36} \\[5ex] \left(x -\dfrac{5}{6}\right)^2 = \dfrac{24}{36} + \dfrac{25}{36} \\[5ex] \left(x -\dfrac{5}{6}\right)^2 = \dfrac{24 + 25}{36} \\[5ex] \left(x -\dfrac{5}{6}\right)^2 = \dfrac{49}{36} \\[5ex] x - \dfrac{5}{6} = \pm \sqrt{\dfrac{49}{36}} \\[5ex] x - \dfrac{5}{6} = \pm \dfrac{7}{6} \\[5ex] x = \dfrac{5}{6} \pm \dfrac{7}{6} \\[5ex] x = \dfrac{5 + 7}{6} \;\;OR\;\; x = \dfrac{5 - 7}{6} \\[5ex] x = \dfrac{12}{6} \;\;OR\;\; x = -\dfrac{2}{6} \\[5ex] x = 2 \;\;OR\;\; x = -\dfrac{1}{3} \\[5ex] \underline{3rd:\;\; Quadratic\;\;Formula} \\[3ex] 3x^2 - 5x = 2 \\[3ex] 3x^2 - 5x - 2 = 0 \\[3ex] Compare\;\;to\;\; ax^2 + bx + c = 0 \\[3ex] a = 3 \\[3ex] b = -5 \\[3ex] c = -2 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} \\[5ex] x = \dfrac{5 \pm \sqrt{25 + 24}}{6} \\[5ex] x = \dfrac{5 \pm \sqrt{49}}{6} \\[5ex] x = \dfrac{5 \pm 7}{6} \\[5ex] x = \dfrac{5 + 7}{6} \;\;OR\;\; x = \dfrac{5 - 7}{6} \\[5ex] x = \dfrac{12}{6} \;\;OR\;\; x = -\dfrac{2}{6} \\[5ex] x = 2 \;\;OR\;\; x = -\dfrac{1}{3} \\[5ex] $ Check

LHS $ 3x^2 - 5x \\[3ex] x = 2 \\[3ex] 3(2)^2 - 5(2) \\[3ex] 3(4) - 10 \\[3ex] 12 - 10 \\[3ex] 2 \\[3ex] $ $ x = -\dfrac{1}{3} \\[5ex] 3\left(-\dfrac{1}{3}\right)^2 - 5\left(-\dfrac{1}{3}\right) \\[5ex] 3\left(\dfrac{1}{9}\right) + \dfrac{5}{3} \\[5ex] \dfrac{1}{3} + \dfrac{5}{3} \\[5ex] \dfrac{1 + 5}{3} \\[5ex] \dfrac{6}{3} \\[5ex] 2 $

RHS $ 2 \\[3ex] $ $ 2 $

Alternative Check
$ 3x^2 - 5x = 2 \\[3ex] 3x^2 - 5x - 2 = 0 \\[3ex] a = 3 \\[3ex] b = -5 \\[3ex] c = -2 \\[3ex] Roots:\;\; x = 2 \;\;OR\;\; x = -\dfrac{1}{3} \\[5ex] $

LHS $ Sum\;\;of\;\;roots \\[5ex] = 2 + -\dfrac{1}{3} \\[5ex] = 2 - \dfrac{1}{3} \\[5ex] = \dfrac{6}{3} - \dfrac{1}{3} \\[5ex] = \dfrac{6 - 1}{3} \\[5ex] = \dfrac{5}{3} \\[5ex] $ $ Product\;\;of\;\;roots \\[3ex] = 2 * -\dfrac{1}{3} \\[5ex] = -\dfrac{2}{3} $

RHS $ Sum\;\;of\;\;roots = \dfrac{-b}{a} \\[5ex] = \dfrac{-(-5)}{3} \\[5ex] = \dfrac{5}{3} \\[5ex] $ $ Product\;\;of\;\;roots = \dfrac{c}{a} \\[5ex] = -\dfrac{2}{3} $

$kx^2 - 23x + 51 = 0$; one solution is $3$

This means that $x = 3$

To find $k$, substitute $x = 3$ into the equation
$ k * (3)^2 - 23(3) + 51 = 0 \\[3ex] k * 9 - 69 + 51 = 0 \\[3ex] 9k = 0 + 69 - 51 \\[3ex] 9k = 18 \\[3ex] k = \dfrac{18}{9} \\[5ex] k = 2 \\[3ex] So,\:\: 2x^2 - 23x + 51 = 0 \\[3ex] $ One solution is a rational number $x = 3$

This implies that the other solution would also be a rational number

You can verify by finding the discriminant.

$ 2x^2 - 23x + 51 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 2 \\[3ex] b = -23 \\[3ex] c = 51 \\[3ex] b^2 - 4ac = (-23)^2 - 4(2)(51) \\[3ex] b^2 - 4ac = 529 - 408 \\[3ex] b^2 - 4ac = 121 = perfect\:\: square \\[3ex] two\:\: rational\:\: roots \\[3ex] (2x^2)(51) = 102x^2 \\[3ex] Factors\:\: are\:\: -17x \:\:and\:\: -6x \\[3ex] 2x^2 - 17x - 6x + 51 = 0 \\[3ex] 2x^2 - 17x = x(2x - 17) \\[3ex] -6x + 51 = -3(2x - 17) \\[3ex] 2x - 17 = 0 \:\:OR\:\: x - 3 = 0 \\[3ex] 2x - 17 = 0 \\[3ex] 2x = 17 \\[3ex] x = \dfrac{17}{2} \\[5ex] x - 3 = 0 \\[3ex] x = 3 \\[3ex] x = \dfrac{17}{2} \:\:OR\:\: x = 3 \\[3ex] $ The other solution is $x = \dfrac{17}{2}$

$x^2 - kx + 13 = 0$; one solution is $3 + 2i$

This means that $x = 3 + 2i$

To find $k$, substitute $x = 3 + 2i$ into the equation

$ (3 + 2i)^2 - k(3 + 2i) + 13 = 0 \\[3ex] (3 + 2i)^2 = (3 + 2i)(3 + 2i) \\[3ex] = 9 + 6i + 6i + 4i^2 \\[3ex] = 9 + 12i + 4(-1) \\[3ex] = 9 + 12i - 4 \\[3ex] = 5 + 12i \\[3ex] \implies (5 + 12i) - 3k - 2ki + 13 = 0 \\[3ex] 5 + 12i - 3k - 2ki + 13 = 0 \\[3ex] 18 + 12i = 3k + 2ki \\[3ex] Equate\:\: corresponding\:\: terms \\[3ex] \implies 3k = 18 \:\:and\:\: 2k = 12 \\[3ex] k = \dfrac{18}{3} \:\:and\:\: k = \dfrac{12}{2} \\[5ex] k = 6 \\[3ex] \implies x^2 - 6x + 13 = 0 \\[3ex] $ One solution is a complex number $x = 3 + 2i$
This implies that the other solution would also be a complex number
You can verify by finding the discriminant.

$ x^2 - 6x + 13 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -6 \\[3ex] c = 13 \\[3ex] b^2 - 4ac = (-6)^2 - 4(1)(13) \\[3ex] = 36 - 52 \\[3ex] = -16 \\[3ex] Discriminant\:\: is\:\: negative \\[3ex] two\:\: complex\:\: roots \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)} \\[5ex] x = \dfrac{6 \pm \sqrt{36 - 52}}{2} \\[5ex] x = \dfrac{6 \pm \sqrt{-16}}{2} \\[5ex] x = \dfrac{6 \pm \sqrt{-1 * 16}}{2} \\[5ex] x = \dfrac{6 \pm \sqrt{-1} * \sqrt{16}}{2} \\[5ex] x = \dfrac{6 \pm i * 4}{2} \\[5ex] x = \dfrac{6 \pm 4i}{2} \\[5ex] x = \dfrac{2(3 \pm 2i)}{2} \\[5ex] x = 3 \pm 2i \\[3ex] x = 3 + 2i \:\:\:OR\:\:\: x = 3 - 2i \\[3ex] $ The other solution is $x = 3 - 2i$

$ For: \\[3ex] ax^2 + bx + c = 0 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] \underline{Compare}: \\[3ex] px^2 + qx + b = 0 \\[3ex] a = p \\[3ex] b = q \\[3ex] c = b \\[3ex] x = x \\[3ex] \implies \\[3ex] x = \dfrac{-q \pm \sqrt{q^2 - 4pb}}{2p} $

$ 2x^2 - 3x - 20 = 0 \\[3ex] 2x^2 * -20 = -40x^2 \\[3ex] Factors\;\;are\;\;5x\;\;and\;\;-8x \\[3ex] 2x^2 + 5x - 8x - 20 = 0 \\[3ex] x(2x + 5) - 4(2x + 5) = 0 \\[3ex] 2x + 5 = 0 \;\;OR\;\; x - 4 = 0 \\[3ex] 2x = -5 \;\;OR\;\; x = 4 \\[3ex] x = -\dfrac{5}{2} \;\;OR\;\; x = 4 $

$ (t - 3)(t + 2) = 0 \\[3ex] Zero\;\;Product\;\;Property \\[3ex] t - 3 = 0 \;\;\;OR\;\;\; t + 2 = 0 \\[3ex] t = 3 \;\;\;OR\;\;\; t = -2 \\[3ex] $ Option B

$ (a) \\[3ex] x^2 - 4y^2 = 0 ...eqn.(1) \\[3ex] x + 2y = 6 ...eqn.(2) \\[3ex] From\;\;eqn.(2) \\[3ex] x = 6 - 2y ...eqn.(3) \\[3ex] Substitute\;\;(6 - 2y)\;\;for\;\;x\;\;in\;\;eqn.(1) \\[3ex] (6 - 2y)^2 - 4y^2 = 0 \\[3ex] (6 - 2y)(6 - 2y) - 4y^2 = 0 \\[3ex] 36 - 12y - 12y + 4y^2 - 4y^2 = 0 \\[3ex] 36 - 24y = 0 \\[3ex] 36 = 24y \\[3ex] 24y = 36 \\[3ex] y = \dfrac{36}{24} \\[5ex] y = \dfrac{3}{2} \\[5ex] Substitute\;\;y = \dfrac{3}{2}\;\;in\;\;eqn.(3) \\[5ex] x = 6 - 2\left(\dfrac{3}{2}\right) \\[5ex] x = 6 - 3 \\[3ex] x = 3 \\[3ex] (x, y) = \left(3, \dfrac{3}{2}\right) \\[5ex] $ Check

$ (x, y) = \left(3, \dfrac{3}{2}\right) $

LHS	RHS
$ x^2 - 4y^2 \\[3ex] 3^2 - 4\left(\dfrac{3}{2}\right)^2 \\[5ex] 9 - 4\left(\dfrac{9}{4}\right) \\[5ex] 9 - 9 \\[3ex] 0 $ $ x + 2y \\[3ex] 3 + 2\left(\dfrac{3}{2}\right) \\[5ex] 3 + 3 \\[3ex] 6 $	$0$ $6$

$ (b) \\[3ex] 1st\;\;Solution:\;\; x = -2 \\[3ex] 1st\;\;Factor:\;\; x + 2 \\[3ex] 2nd\;\;Solution:\;\; x = 3 \\[3ex] 2nd\;\;Factor:\;\; x - 3 \\[3ex] Factors = (x + 2)(x - 3) \\[3ex] \implies \\[3ex] x^2 - 3x + 2x - 6 \\[3ex] x^2 - x - 6 $

$ Roots:\;\; \dfrac{1}{2} \;\;and\;\; -3 \\[5ex] Factors:\;\; \left(x - \dfrac{1}{2}\right) \;\;and\;\; (x + 3) \\[5ex] \implies \\[3ex] \left(x - \dfrac{1}{2}\right)(x + 3) = 0 \\[5ex] x^2 + 3x - \dfrac{1}{2}x - \dfrac{3}{2} = 0 \\[5ex] x^2 + \dfrac{6x - x}{2} - \dfrac{3}{2} = 0 \\[5ex] x^2 + \dfrac{5}{2}x - \dfrac{3}{2} = 0 \\[5ex] LCD = 2 \\[3ex] 2(x^2 + \dfrac{5}{2}x - \dfrac{3}{2}) = 2(0) \\[5ex] 2x^2 + 5x - 3 = 0 \\[3ex] Compare: \\[3ex] 2x^2 + 5x + R = 0 \\[3ex] R = -3 $

We can solve this in two ways.
First Method: Alternative Check (Sum and Product of Roots)
This method is recommended because the ACT is a timed test.
$ ax^2 + bx + c = 0 \\[3ex] x^2 + x - 30 = 0 \\[3ex] a = 1 \\[3ex] b = 1 \\[3ex] c = -30 \\[3ex] Sum\:\: of\:\: roots = -\dfrac{b}{a} \\[5ex] = -\dfrac{1}{1} \\[3ex] = -1 \\[3ex] $ Second Method: Factoring Method
$ x^2 + x - 30 = 0 \\[3ex] \Rightarrow (x + 6)(x - 5) = 0 \\[3ex] x + 6 = 0 \\[3ex] x = -6 \\[3ex] OR \\[3ex] x - 5 = 0 \\[3ex] x = 5 \\[3ex] \Rightarrow x = -6 \:\:\:OR\:\:\: x = 5 \\[3ex] Sum\:\: of\:\: roots = -6 + 5 = -1 $