Question

What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3? x = –2 – StartRoot StartFraction 11 Over 2 EndFraction EndRoot and x = –2 + StartRoot StartFraction 11 Over 2 EndFraction EndRoot x = –2 – StartRoot StartFraction 7 Over 2 EndFraction EndRoot and x = –2 + StartRoot StartFraction 7 Over 2 EndFraction EndRoot x = 2 – StartRoot StartFraction 11 Over 2 EndFraction EndRoot and x = 2 + StartRoot StartFraction 11 Over 2 EndFraction EndRoot x = 2 – StartRoot StartFraction 7 Over 2 EndFraction EndRoot and x = 2 + StartRoot StartFraction 7 Over 2 EndFraction EndRoot

Answer 1

Answer:

The zeros are

$x=-2+\frac{\sqrt{22}}{2}$

$x=-2-\frac{\sqrt{22}}{2}$  

Step-by-step explanation:

we have

$f(x)=2x^{2}+8x-3$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

Equate the function to zero

$2x^{2} +8x-3=0$  

so

$a=2\\b=8\\c=-3$

substitute in the formula      

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

$x=\frac{-8\pm\sqrt{88}} {4}$ 

$x=\frac{-8\pm2\sqrt{22}} {4}$ 

$x=-2\pm\frac{\sqrt{22}}{2}$

therefore

$x=-2+\frac{\sqrt{22}}{2}$

$x=-2-\frac{\sqrt{22}}{2}$  

Answer 2

Answer:

correct answer is d

Step-by-step explanation: