Question

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3? x = StartFraction negative 5 Over 2 EndFraction – StartFraction StartRoot 31 EndRoot Over 2 EndFraction and x = StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 31 EndRoot Over 2 EndFraction x = StartFraction negative 5 Over 2 EndFraction – StartFraction StartRoot 37 EndRoot Over 8 EndFraction and x = StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 37 EndRoot Over 8 EndFraction x = StartFraction 5 Over 2 EndFraction – StartFraction StartRoot 31 EndRoot Over 2 EndFraction and x = StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 31 EndRoot Over 2 EndFraction x = StartFraction 5 Over 2 EndFraction – StartFraction StartRoot 37 EndRoot Over 8 EndFraction and x = StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 37 EndRoot Over 8 EndFraction

Answer 1

Answer:

$x=\frac{5}{2}+\frac{\sqrt{31}}{2}\ \ or\ \ x=\frac{5}{2}-\frac{\sqrt{31}}{2}$

Step-by-step explanation:

Given the following Quadratic function:

$f(x) = 2x^2 - 10x -3$

We must make it equal to zero:

$0 = 2x^2 - 10x -3$

Now we need to apply the Quadratic formula. This is:

$x=\frac{-b\± \sqrt{b^2-4ac}}{2a}$

In this case we can identify that:

$a=2\\b=-10\\c=-3$

Finally, substituting these values into the Quadratic formula, we get the following solutions:

$x=\frac{-(-10)\±\sqrt{(-10)^2-4(2)(-3)}}{2(2)}$

$x=\frac{5}{2}+\frac{\sqrt{31}}{2}\ \ or\ \ x=\frac{5}{2}-\frac{\sqrt{31}}{2}$

Answer 2

Answer:

Step-by-step explanation:

It’s D