Solving Equations Using the Quadratic Formula

Question 1 of 6

Solve using the quadratic formula

$x^2+2x-24=0$

1. $-4$ and $6$
2. $-4$ and $-6$
3. $4$ and $-6$
4. $4$ and $6$

Incorrect

The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, list the coefficients of the quadratic equation individually

$x^2+2x-24=0$

$a=1$ $b=2$ $c=-24$

Substitute the values into the Quadratic Formula

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{2} \pm \sqrt {\color{#9a00c7}{2}^2-4\color{#00880A}{(1)}\color{#007DDC}{(-24)}} }{2 \color{#00880A}{(1)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {-2 \pm \sqrt {4 +96} }{2}$

	$=$	$\frac {-2 \pm \sqrt {100} }{2}$

	$=$	$\frac {-2 \pm 10 }{2}$

Write each root individually

$x_1$	$=$	$\frac {-2 + 10 }{2}$

	$=$	$\frac {8}{2}$

	$=$	$4$

$x_2$	$=$	$\frac {-2 – 10 }{2}$

	$=$	$\frac {-12}{2}$

	$=$	$-6$

$x = 4,-6$

Question 2 of 6

Solve using the quadratic formula

$8x^2-8x-3=0$

1. $-1.29$ and $0.29$
2. $1.29$ and $-0.29$
3. $-1.29$ and $-0.29$
4. $1.29$ and $0.29$

Incorrect

The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, list the coefficients of the quadratic equation individually

$8x^2-8x-3=0$

$a=8$ $b=-8$ $c=-3$

Substitute the values into the Quadratic Formula

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{-8} \pm \sqrt {\color{#9a00c7}{-8}^2-4\color{#00880A}{(8)}\color{#007DDC}{(-3)}} }{2 \color{#00880A}{(8)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {8 \pm \sqrt {64 +96} }{16}$

	$=$	$\frac {8 \pm \sqrt {160} }{16}$

	$=$	$\frac {8 \pm 4 \sqrt{10} }{16}$

	$=$	$\frac {2 \pm \sqrt{10} }{4}$	Simplify

Write each root individually

$x_1$	$=$	$\frac {2 + \sqrt{10} }{4}$

	$=$	$1.29$

$x_2$	$=$	$\frac {2-\sqrt{10} }{4}$

	$=$	$-0.29$

$x = 1.29,-0.29$

Question 3 of 6

Solve using the quadratic formula

$x^2+11x+20=0$

1. $-2.298$ and $-8.702$
2. $2.298$ and $8.702$
3. $-2.298$ and $8.702$
4. $2.298$ and $-8.702$

Incorrect

The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, list the coefficients of the quadratic equation individually

$x^2+11x+20=0$

$a=1$ $b=11$ $c=20$

Substitute the values into the Quadratic Formula

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{11} \pm \sqrt {\color{#9a00c7}{11}^2-4\color{#00880A}{(1)}\color{#007DDC}{(20)}} }{2 \color{#00880A}{(1)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {-11 \pm \sqrt {121 -80} }{2}$

	$=$	$\frac {-11 \pm \sqrt {41} }{2}$

Write each root individually

$x_1$	$=$	$\frac {-11 + \sqrt {41} }{2}$

	$=$	$-2.298$

$x_2$	$=$	$\frac {-11-\sqrt {41} }{2}$

	$=$	$-8.702$

$x = -2.298,-8.702$

Question 4 of 6

Solve using the quadratic formula

$-x^2-3x=-9$

1. $4.854$ and $-1.854$
2. $-4.854$ and $-1.854$
3. $4.854$ and $1.854$
4. $-4.854$ and $1.854$

Incorrect

The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, convert the equation to standard form

$-x^2-3x$ $+9$	$=$	$-9$ $+9$
$-x^2-3x+9$	$=$	$0$

First, list the coefficients of the quadratic equation individually

$-x^2-3x+9=0$

$a=-1$ $b=-3$ $c=9$

Substitute the values into the Quadratic Formula

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{-3} \pm \sqrt {\color{#9a00c7}{-3}^2-4\color{#00880A}{(-1)}\color{#007DDC}{(9)}} }{2 \color{#00880A}{(-1)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {3 \pm \sqrt {9 +36} }{-2}$

	$=$	$\frac {3 \pm \sqrt {45} }{-2}$

	$=$	$\frac {3 \pm 3\sqrt{5} }{-2}$

Write each root individually

$x_1$	$=$	$\frac {3 + 3\sqrt{5} }{-2}$

	$=$	$-4.854$

$x_2$	$=$	$\frac {3-3\sqrt{5} }{-2}$

	$=$	$1.854$

$x = -4.854, 1.854$

Question 5 of 6

Solve using the quadratic formula

$x-3/x=4$

1. $2+sqrt7$ and $2+sqrt7$
2. $2+sqrt7$ and $2-sqrt7$
3. $-2+sqrt7$ and $-2-sqrt7$
4. $2-sqrt7$ and $2-sqrt7$

Incorrect

The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, convert the equation to standard form

$x-3/x$ $timesx$	$=$	$4$ $timesx$
$x^2-3$ $-4x$	$=$	$4x$ $-4x$
$x^2-4x-3$	$=$	$0$

First, list the coefficients of the quadratic equation individually

$x^2-4x-3=0$

$a=1$ $b=-4$ $c=-3$

Substitute the values into the Quadratic Formula

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{-4} \pm \sqrt {\color{#9a00c7}{-4}^2-4\color{#00880A}{(1)}\color{#007DDC}{(-3)}} }{2 \color{#00880A}{(1)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {4 \pm \sqrt {16+12} }{2}$

	$=$	$\frac {4 \pm \sqrt {28} }{2}$

	$=$	$\frac {4 \pm 2\sqrt{7} }{2}$

	$=$	$2 \pm \sqrt{7}$	Simplify

$x = 2+\sqrt7, 2-\sqrt7$

Question 6 of 6

Solve using the quadratic formula

$4(x+5)^2=48$

1. $-5+-2sqrt3$
2. $5+-3sqrt2$
3. $5+-2sqrt3$
4. $-5+-3sqrt2$

Incorrect

The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, write the given equation in standard form ( $a$

$x^2+$ $b$

$x+$ $c$ )

$4(x+5)^2$	$=$	$48$
$4(x+5)^2$ $-:4$	$=$	$48$ $-:4$	Divide both sides by $4$
$(x+5)^2$	$=$	$12$
$x^2+10x+25$	$=$	$12$
$x^2+10x+25$ $-12$	$=$	$12$ $-12$	Subtract $12$ from both sides
$x^2+10x+13$	$=$	$0$

Next, list the coefficients of the quadratic equation individually

$x^2+10x+13=0$

$a=1$ $b=10$ $c=13$

Substitute the values into the Quadratic Formula

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{10} \pm \sqrt {\color{#9a00c7}{10}^2-4\color{#00880A}{(1)}\color{#007DDC}{(13)}} }{2 \color{#00880A}{(1)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {-10 \pm \sqrt {100 -52} }{2}$

	$=$	$\frac {-10 \pm \sqrt {48} }{2}$

	$=$	$\frac {-10 \pm 4\sqrt3 }{2}$

	$=$	$-5 \pm 2\sqrt3$

The roots can also be written individually

$=$

$-5+2\sqrt3$

$=$

$-5- 2\sqrt3$

$x = -5 \pm 2\sqrt3$

Solving Equations Using the Quadratic Formula

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The Quadratic Formula

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The Quadratic Formula

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The Quadratic Formula

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The Quadratic Formula

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