Question

unit 5 review mr. morgan directions: answer the following questions. if you used a calculator for questions which require using the quadratic formula, you must show your values for a,b, and c as well as show where you your values into the quadratic formula. all questions must be attempted and have work if y receive full credit. decimal answers will not be accepted on the review nor the exam. exercise #1: solve for the roots by factoring (1) $x^2 - 13x + 42 = 0$ (2) $x^2 + 10x + 16 = 0$ (3) $2x^2 - 2x - 24 = 0$ (4) $4x^2 + 15x + 9 = 0$ exercise #2: solve by using the quadratic formula. answer must be in simplest radical form (5) $x^2 + 3x + 1 = 0$ (6) $4x^2 - x - 11 = 0$ (7) $2x^2 - 8 = 3x$ (8) $x^2 + 6x = 4$

Explanation:

Exercise #1: Solve by Factoring

(1) Step1: Find factor pairs of 42

We need two numbers that multiply to $42$ and add to $-13$. These numbers are $-6$ and $-7$.

(1) Step2: Factor the quadratic

$x^2 -13x +42=(x-6)(x-7)=0$

(1) Step3: Solve for $x$

Set each factor equal to 0:
$x-6=0 \implies x=6$
$x-7=0 \implies x=7$

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(2) Step1: Find factor pairs of 16

We need two numbers that multiply to $16$ and add to $10$. These numbers are $2$ and $8$.

(2) Step2: Factor the quadratic

$x^2 +10x +16=(x+2)(x+8)=0$

(2) Step3: Solve for $x$

Set each factor equal to 0:
$x+2=0 \implies x=-2$
$x+8=0 \implies x=-8$

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(3) Step1: Factor out GCF first

The greatest common factor of $2x^2-2x-24$ is $2$:
$2(x^2 -x -12)=0$

(3) Step2: Factor the quadratic inside

Find two numbers that multiply to $-12$ and add to $-1$: $-4$ and $3$.
$x^2 -x -12=(x-4)(x+3)$

(3) Step3: Solve for $x$

$2(x-4)(x+3)=0$
$x-4=0 \implies x=4$
$x+3=0 \implies x=-3$

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(4) Step1: Factor the quadratic

We need two numbers that multiply to $4\times9=36$ and add to $15$. These numbers are $12$ and $3$. Split the middle term:
$4x^2 +12x +3x +9=0$

(4) Step2: Factor by grouping

$4x(x+3)+3(x+3)=0 \implies (4x+3)(x+3)=0$

(4) Step3: Solve for $x$

$4x+3=0 \implies x=-\frac{3}{4}$
$x+3=0 \implies x=-3$

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Exercise #2: Solve with Quadratic Formula

Quadratic formula: For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(5) Step1: Identify $a,b,c$

For $x^2+3x+1=0$, $a=1$, $b=3$, $c=1$

(5) Step2: Substitute into formula

$x=\frac{-3\pm\sqrt{3^2-4(1)(1)}}{2(1)}=\frac{-3\pm\sqrt{9-4}}{2}=\frac{-3\pm\sqrt{5}}{2}$

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(6) Step1: Identify $a,b,c$

For $4x^2-x-11=0$, $a=4$, $b=-1$, $c=-11$

(6) Step2: Substitute into formula

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(4)(-11)}}{2(4)}=\frac{1\pm\sqrt{1+176}}{8}=\frac{1\pm\sqrt{177}}{8}$

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(7) Step1: Rewrite in standard form

$2x^2-3x-8=0$, so $a=2$, $b=-3$, $c=-8$

(7) Step2: Substitute into formula

$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-8)}}{2(2)}=\frac{3\pm\sqrt{9+64}}{4}=\frac{3\pm\sqrt{73}}{4}$

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(8) Step1: Rewrite in standard form

$x^2+6x-4=0$, so $a=1$, $b=6$, $c=-4$

(8) Step2: Substitute into formula

$x=\frac{-6\pm\sqrt{6^2-4(1)(-4)}}{2(1)}=\frac{-6\pm\sqrt{36+16}}{2}=\frac{-6\pm\sqrt{52}}{2}=\frac{-6\pm2\sqrt{13}}{2}=-3\pm\sqrt{13}$

Answer:

Exercise #1:

1. $x=6$ and $x=7$
2. $x=-2$ and $x=-8$
3. $x=4$ and $x=-3$
4. $x=-\frac{3}{4}$ and $x=-3$

Exercise #2:

1. $x=\frac{-3+\sqrt{5}}{2}$ and $x=\frac{-3-\sqrt{5}}{2}$
2. $x=\frac{1+\sqrt{177}}{8}$ and $x=\frac{1-\sqrt{177}}{8}$
3. $x=\frac{3+\sqrt{73}}{4}$ and $x=\frac{3-\sqrt{73}}{4}$
4. $x=-3+\sqrt{13}$ and $x=-3-\sqrt{13}$