Respuesta :
Answer:
1) [tex] x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}[/tex]
[tex] x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}[/tex]
2) [tex] x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}[/tex]
[tex] x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}[/tex]
3) [tex] p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}[/tex]
[tex] p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}[/tex]
4) [tex] y_1 = \frac{-3 - 9}{4}=-3 [/tex]
[tex] y_2 = \frac{-3 + 9}{4}= \frac{3}{2}[/tex]
Step-by-step explanation:
The quadratic formula is given by:
[tex] x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
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:
[tex] x = \frac{-(-2) \pm \sqrt{(-2)^2 -4(1)(-12)}}{2(1)}[/tex]
[tex] x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}[/tex]
[tex] x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}[/tex]
2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2
For this case if we apply the quadratic formula we got:
[tex] x = \frac{-(-6) \pm \sqrt{(-6)^2 -4(1/2)(-2)}}{2(1/2)}[/tex]
[tex] x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}[/tex]
[tex] x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}[/tex]
3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7
For this case if we apply the quadratic formula we got:
[tex] p = \frac{-(8) \pm \sqrt{(8)^2 -4(2)(-7)}}{2(2)}[/tex]
[tex] p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}[/tex]
[tex] p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}[/tex]
4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9
For this case if we apply the quadratic formula we got:
[tex] y = \frac{-(3) \pm \sqrt{(3)^2 -4(2)(-9)}}{2(2)}[/tex]
[tex] y_1 = \frac{-3 - 9}{4}=-3 [/tex]
[tex] y_2 = \frac{-3 + 9}{4}= \frac{3}{2}[/tex]
