Question

The function, f(x) = –2x2 + x + 5, is in standard form. The quadratic equation is 0 = –2x2 + x + 5, where a = –2, b = 1, and c = 5. The discriminate b2 – 4ac is 41. Now, complete step 5 to solve for the zeros of the quadratic function. 5. Solve using the quadratic formula. x = What are the zeros of the function f(x) = x + 5 – 2x2? x = x = x = x =

Answer 1

Given:

The function, f(x) = -2x^2 + x + 5

Quadratic equation: 0 = -2x^2 + x +5
where a = -2
            b = 1
            c = 5

The discriminate b^2 - 4ac = 41

To solve for the zeros of the quadratic function, use this formula:

x = ( -b +-√ (b^2 - 4ac) ) / 2a

x = ( 1 + √41 ) / 4  or 1.85
x = ( 1 - √41 ) / 4  or -1.35

Therefore, the zeros of the quadratic equation are 1.85 and -1.35. 

Answer 2

Answer:

A quadratic equation $ax^2+bx+c=0$   ....[1] where, a, b and c are coefficient then;

the solution fore this equation we have;

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

Given the function in the standard form:

$f(x) = -2x^2+x+5$

The quadratic equation is:

$0 = -2x^2+x+5$

On comparing with equation [1] we have;

a = -2 , b = 1 and c = 5.

The discriminant is given by:

$\text{Discriminant} = b^2-4ac$

then;

$\text{Discriminant} = (1)^2-4(-2)(5) = 1+40 = 41$

Solve for the zeros of the quadratic function:

Substitute the given values in [1] we have;

$x = \frac{-1 \pm \sqrt{41}}{2(-2)} = \frac{-1 \pm \sqrt{41}}{-4}$

then;

$x = \frac{-1 + \sqrt{41}}{-4}$ and $x = \frac{-1 - \sqrt{41}}{-4}$

⇒$x = \frac{1 - \sqrt{41}}{4}$ and $x = \frac{1 + \sqrt{41}}{4}$

Therefore, the zeros of the given  function are:

$x = \frac{1 - \sqrt{41}}{4}$ , $\frac{1 + \sqrt{41}}{4}$