Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples - Quadratic Equations

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve and check each equation if asked to use any specific method.
If you used multiple methods to solve an equation, check your solution at least one time.
For each question, use the Main Check ($LHS = RHS$) for the Factoring method or Completing the Square method.
For each question for which you checked already with the Main Check ($LHS = RHS$), use the Alternative Check (Sum and Product of Roots) for the Quadratic Formula method.
Solve and "reasonably" check all word problems.
Use at least three methods for each equation as applicable.
For each equation, calculate the discriminant only one time.
State the nature of the roots of the equation based on the discriminant.
Indicate the method used for solving each equation.
Show all Work.
Integers and fractions are only allowed for each equation.

(1.) $x^2 - 6x - 40 = 0$
Use the Factoring method if applicable.


$ x^2 - 6x - 40 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1, b = -6, 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 $

(2.) $x^2 - 6x - 40 = 0$
Use the Completing the Square method.


$ 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 $
(3.) $x^2 - 6x - 40 = 0$
Use the Quadratic Formula.


$ x^2 - 6x - 40 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 1, b = -6, 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 = $10 + (-4) = 10 - 4 = 6$


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

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

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


$ \dfrac{c}{a} = \dfrac{-40}{1} = -40 $
(4.) $-x^2 + 2x = 1$
Use the Factoring method if applicable.


$ -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, b = -2, 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 $

(5.) $-x^2 + 2x = 1$
Use the Completing the Square method.


$ -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) $
(6.) $-x^2 + 2x = 1$
Use the Quadratic Formula.


$ -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, b = -2, 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 = $1 + 1 = 2$


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

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

Product of roots = $1 * 1 = 1$


$ \dfrac{c}{a} = \dfrac{1}{1} = 1 $
(7.) $-10 = 17x - 20x^2$
Use the Factoring method if applicable.


$ -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, b = -17, 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 $
(8.) $-10 = 17x - 20x^2$
Use the Completing the Square method.


$ -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} $
(9.) $-10 = 17x - 20x^2$
Use the Quadratic Formula.


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

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] $
$ \dfrac{c}{a} = \dfrac{-10}{20} \\[5ex] = -\dfrac{1}{2} $
(10.) For the equation: $-px + 35x^2 = -3$, list all the values of $p$ for which the equation can be solved by Factoring.


$ -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] $
(11.) ACT What is the sum of the $2$ solutions of the equation
$x^2 + x - 30 = 0$?


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, b = 1, 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 $
(12.) ACT A function $f(x)$ is defined as $f(x) = 3^{x^2 - x - 2}$. What $2$ real numbers satisfy $f(x) = 1$?



$ 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] 3^{x^2 - x - 2} = 3^0 \\[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] $
(13.) ACT Which of the following most precisely describes the roots of the equation: $5x^2 + 7x + 2 = 0$?
F. $1$ rational (double) root
G. $1$ irrational (double) root
H. $2$ rational roots
J. $2$ irrational roots
K. $2$ complex roots (with nonzero imaginary parts)


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, b = 7, 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.
(14.) ACT One solution of the equation $4x^3 - 2x^2 + x + 7 = 0$ is $x = -1$. Which of the following describes the other $2$ solutions?
F. Both are negative real numbers.
G. One is a negative real number, and the other is a positive real number.
H. Both are positive real numbers.
J. One is a positive real number, and the other is a complex number that is not real.
K. Both are complex numbers that are not real.


$4x^3 - 2x^2 + x + 7 = 0$
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, b = -6, 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.
(15.) $5x^2 + 6x = -9$
Use the Factoring method if applicable.
If it is not applicable, use the Completing the Square method.


$ 5x^2 + 6x = -9 \\[3ex] 5x^2 + 6x + 9 = 0 \\[3ex] ax^2 + bx + c = 0 \\[3ex] a = 5, b = 6, 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 $

(16.) $5x^2 + 6x = -9$
Use the Quadratic Formula.


$ ax^2 + bx + c = 0 \\[3ex] 5x^2 + 6x = -9 \\[3ex] 5x^2 + 6x + 9 = 0 \\[3ex] a = 5, b = 6, 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, b = 6, 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] $
$\dfrac{-b}{a} = \dfrac{-6}{5} \\[5ex]$

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] $
$\dfrac{c}{a} = \dfrac{9}{5}$
(17.) For the following equation under the given condition,
(a.) Find $k$ and
(b.) Find a second solution

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


$kx^2 - 23x + 51 = 0$; 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, b = -23, 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 $x = \dfrac{17}{2}$
(18.) For the following equation under the given condition,
(a.) Find $k$ and
(b.) Find a second solution

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


$x^2 - kx + 13 = 0$; 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, b = -6, 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 $x = 3 - 2i$
(19.) ACT For what integer are both solutions of the equation $x^2 + kx + 17 = 0$ positive integers?


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 $
(20.) ACT The solution set for the equation $2^{x^2 + 1} = 1$ contains:

F. $2$ imaginary numbers.
G. $2$ positive real numbers.
H. $1$ negative and $1$ positive real number.
J. $1$ negative real number only.
K. $1$ real number, which is $0$.


$ 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




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(21.) ACT If $x^2 + 4 = 29$, then $x^2 - 4 = ?$

$ A.\:\: 5 \\[3ex] B.\:\: \sqrt{21} \\[3ex] C.\:\: 21 \\[3ex] D.\:\: 25 \\[3ex] E.\:\: 33 \\[3ex] $

$ x^2 + 4 = 29 \\[3ex] x^2 = 29 - 4 \\[3ex] x^2 = 25 \\[3ex] x^2 - 4 = 25 - 4 = 21 $
(22.) ACT What is the solution set of the equation $4x^2 - 9 = 0?$

$ A.\:\: \{-\sqrt{2}, \sqrt{2}\} \\[3ex] B.\:\: \left\{-\dfrac{3}{2}, \dfrac{3}{2}\right\} \\[5ex] C.\:\: \{-\sqrt{3}, \sqrt{3}\} \\[3ex] D.\:\: \{-2, 2\} \\[3ex] E.\:\: \left\{\dfrac{9}{4}\right\} \\[5ex] $

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} $
(23.) JAMB Find the values of $p$ for which the equation $x^2 - (p - 2)x + 2p + 1 = 0$ has equal roots.

$ A.\:\: (0, 12) \\[3ex] B.\:\: (1, 2) \\[3ex] C.\:\: (21, 0) \\[3ex] D.\:\: (4, 5) \\[3ex] E.\:\: (3, 4) \\[3ex] $

$ 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 $
Determine whether the ordered pairs are solutions of the equations.
(24.) $-x^2 - y^2 = -29$
Ordered pair = $(2, -5)$


$(2, -5)$ 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. $