Question

Match each quadratic equation with its solution set.

Answer 1

For this case we solve each of the equations:

$2x ^ 2-32 = 0$

Adding 32 to both sides of the equation:

$2x ^ 2 = 32$

Dividing between 2 on both sides of the equation:

$x ^ 2 = \frac {32} {2}\\x ^ 2 = 16$

Applying square root to eliminate the exponent:

$x = \pm \sqrt {16}\\x = \pm4\\x_ {1} = + 4\\x_ {2} = - 4$

Second equation:

$4x ^ 2-100 = 0$

Adding 100 to both sides of the equation:

$4x ^ 2 = 100$

Dividing between 4 on both sides of the equation:

$x ^ 2 = \frac {100} {4}\\x ^ 2 = 25$

Applying square root to eliminate the exponent:

$x = \pm \sqrt {25}\\x = \pm5\\x_ {1} = + 5\\x_ {2} = - 5$

Third equation:

$x ^ 2-55 = 9$

Adding 55 to both sides of the equation:

$x ^ 2 = 9 + 55\\x ^ 2 = 64$

Applying square root to eliminate the exponent:

$x = \sqrt {64}\\x = \pm8$

Fourth equation:

$x ^ 2-140 = -19$

Adding 140 to both sides of the equation:

$x ^ 2 = -19 + 140\\x ^ 2 = 121$

Applying square root to eliminate the exponent:

$x = \pm \sqrt {121}\\x = \pm11$

Fifth equation:

$2x ^ 2-18 = 0$

Adding 18 to both sides of the equation:

2x ^ 2 = 18

Dividing between 2 on both sides of the equation:

$x ^ 2 = \frac {18} {2}\\x ^ 2 = 9$

Applying square root to eliminate the exponent:

$x = \pm \sqrt {9}\\x = \pm3$

Answer:

$2x ^ 2-32 = 0, x_ {1} = + 4, x_ {2} = - 4\\4x ^ 2-100 = 0, x_ {1} = + 5, x_ {2} = - 5\\x ^ 2-55 = 9, x_ {1} = + 8, x_ {2} = - 8\\x ^ 2-140 = -19, x_ {1} = + 11, x_ {2} = - 11\\2x ^ 2-18 = 0, x_ {1} = + 3, x_ {2} = - 3$

Answer 2

2x2-32=0 is -4,4

4x2-100=0 is -5,5