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Find the zeros of f(x) = x^2 + 7x + 9

1 Answer

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Answer:

-7/2 ±1/2sqrt(13) = x

Explanation:

f(x) =x^2 + 7x + 9

To find the zeros, set this equal to zero

0 = x^2 + 7x + 9

I will complete the square

Subtract 9 from each side

0-9 = x^2 + 7x + 9-9

-9 =x^2 + 7x

Take the coefficient of the x term, 7

divide by 2, 7/2

Then square it, (7/2)^2 = 49/4

Add this to both sides

-9 +49/4=x^2 + 7x + 49/4

-36/4 +49/4 = (x+7/2)^2

13/4 = (x+7/2)^2

Take the square root of each side

±sqrt(13/4) = sqrt( (x+7/2)^2)

± sqrt(13) /sqrt(4)= (x+7/2)

± 1/2 sqrt(13) = (x+7/2)

Subtract 7/2 from each side

-7/2 ±1/2sqrt(13) = x+7/2-7/2

-7/2 ±1/2sqrt(13) = x

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