Question

What are the zeros of the quadratic function f(x) = 8x2 – 16x – 15? x = –1 – Startroot 2 EndRoot and x = –1 + Startroot 2 EndRoot x = –1 – StartRoot StartFraction 15 Over 8 EndFraction EndRoot and x = –1 + StartRoot StartFraction 15 Over 8 EndFraction EndRoot x = 1 – StartRoot StartFraction 23 Over 8 EndFraction EndRoot and x = 1 + StartRoot StartFraction 23 Over 8 EndFraction EndRoot x = 1 – StartRoot 7 EndRoot and x = 1 + StartRoot 7 EndRoot

Answer 1

Answer:

$x =1 + \sqrt{ \frac{23}{8} }$

or

$x =1 - \sqrt{ \frac{23}{8} }$

Step-by-step explanation:

The given quadratic function is

$f(x) = 8 {x}^{2} - 16x - 15$

To find the roots, we set f(x)=0 to get:

$8 {x}^{2} - 16x - 15 = 0$

The solution is given by:

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

where a=8, b=-16, c=-15

We substitute to get:

$x = \frac{ - - 16 \pm \sqrt{ {( -1 6)}^{2} - 4 \times 8 \times - 15} }{2 \times 8}$

We simplify to get:

$x = \frac{ 16 \pm \sqrt{256 + 480} }{16}$

$x =1 \pm \sqrt{ \frac{23}{8} }$

$x =1 - \sqrt{ \frac{23}{8} }$

or

$x =1 + \sqrt{ \frac{23}{8} }$

Answer 2

Answer:

C. on Edge

Got 100% on the test

Step-by-step explanation:

Proof in the attached picture.