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What are the zeros of the function f (x)=x^2+5x+5 written in simplest radical form​

User Ncw
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1 Answer

5 votes

Answer:

The zeros are


x1=\frac{-5+√(5)} {2} and
x2=\frac{-5-√(5)} {2}

Explanation:

we know that

The formula to solve a quadratic equation of the form
ax^(2) +bx+c=0 is equal to


x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}

in this problem we have


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

To find the zeros equate the function to 0


x^(2) +5x+5=0

so


a=1\\b=5\\c=5

substitute in the formula


x=\frac{-5(+/-)\sqrt{5^(2)-4(1)(5)}} {2(1)}


x=\frac{-5(+/-)√(5)} {2}


x1=\frac{-5+√(5)} {2}


x2=\frac{-5-√(5)} {2}

User YaTaras
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