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What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?




Quadratic formula: x = StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction


x = StartFraction 5 plus or minus 10 StartRoot 5 EndRoot Over 2 EndFraction

x = StartFraction negative 5 plus or minus 10 StartRoot 5 EndRoot Over 2 EndFraction

x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

x = StartFraction 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

User Kohi
by
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2 Answers

2 votes

Answer:

The Answer Is C

Explanation:

x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

User IAmNoone
by
4.0k points
2 votes

Answer:

(-5+√5)/2 and (-5-√5)/2

Explanation:

For a general quadratic equation of the form ax²+bx+c = 0

The value of x can be gotten using the general formula expressed as:

x = -b±√b²-4ac/2a

Now given the function

f(x) = x² + 5x + 5

We can get the zeros of the function in its radical form by simply applying the formula above.

From the equation given;

a = 1, b = 5 and c = 5

Substituting this values in the formula we have:

x = -5±√5²-4(1)(5)/2(1)

x = -5±√25-20/2

x = -5±√5/2

x = (-5+√5)/2 and (-5-√5)/2

The zeros of the function in its simplest radical form are

(-5+√5)/2 and (-5-√5)/2

User MrMikimn
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4.8k points