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IrinaK [193]
3 years ago
5

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 =
Mathematics
2 answers:
avanturin [10]3 years ago
4 0
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. 
professor190 [17]3 years ago
3 0

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}

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Answer:

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Upper quartile: 46

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