Question

Use the quadratic formula to solve each equation.

Answer 1

Answer:

1) $x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}$

$x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}$

2) $x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}$

$x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}$

3) $p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}$

$p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}$

4) $y_1 = \frac{-3 - 9}{4}=-3$

$y_2 = \frac{-3 + 9}{4}= \frac{3}{2}$

Step-by-step explanation:

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

We can use this formula in order to solve the following equations:

1. x^2 − 2x = 12 → a = 1, b = −2, c = −12

For this case if we apply the quadratic formula we got:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 -4(1)(-12)}}{2(1)}$

$x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}$

$x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}$

2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2

For this case if we apply the quadratic formula we got:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 -4(1/2)(-2)}}{2(1/2)}$

$x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}$

$x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}$

3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7

For this case if we apply the quadratic formula we got:

$p = \frac{-(8) \pm \sqrt{(8)^2 -4(2)(-7)}}{2(2)}$

$p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}$

$p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}$

4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9

For this case if we apply the quadratic formula we got:

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

$y_1 = \frac{-3 - 9}{4}=-3$

$y_2 = \frac{-3 + 9}{4}= \frac{3}{2}$