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

A

Explanation:

on edge

Answer 2

The zeros are

`x=-2+(√(22))/(2)`

`x=-2-(√(22))/(2)`

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√(88)} {4}`

`x=\frac{-8\pm2√(22)} {4}`

`x=-2\pm(√(22))/(2)`

therefore

`x=-2+(√(22))/(2)`

`x=-2-(√(22))/(2)`