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andrey2020 [161]
3 years ago
12

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

ative 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

Mathematics
2 answers:
Harman [31]3 years ago
6 0

Answer:

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

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

Step-by-step 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  

equate the function to zero

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(+/-)\sqrt{5}} {2}

therefore

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

frutty [35]3 years ago
6 0

The zeros of the function f(x) are :

<h3>x = ( -5 ± √5 ) / 2</h3>

<h3>Further explanation</h3>

Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :

<h2>D = b² - 4 a c</h2>

From the value of Discriminant , we know how many solutions the equation has by condition :

D < 0 → No Real Roots

D = 0 → One Real Root

D > 0 → Two Real Roots

An axis of symmetry of quadratic equation y = ax² + bx + c is :

\large {\boxed {x = \frac{-b}{2a} } }

Let us now tackle the problem!

<u>Given:</u>

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

<em>The zeros of the quadratic function could be calculated when</em> f(x) = 0 :

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

0 = x^2 + 5x + 5

0 = (x + \frac{5}{2})^2 - (\frac{5}{2})^2 + 5

(x + \frac{5}{2})^2 = (\frac{5}{2})^2 - 5

(x + \frac{5}{2})^2 = \frac{25}{4} - 5

(x + \frac{5}{2})^2 = \frac{5}{4}

x + \frac{5}{2} = \pm \sqrt{\frac{5}{4}}

x + \frac{5}{2} = \pm \frac{\sqrt{5}}{2}

x = - \frac{5}{2} \pm \frac{\sqrt{5}}{2}

x = \boxed{ \frac{-5 \pm \sqrt{5}}{2} }

\texttt{ }

<h3>Learn more</h3>
  • Solving Quadratic Equations by Factoring : brainly.com/question/12182022
  • Determine the Discriminant : brainly.com/question/4600943
  • Formula of Quadratic Equations : brainly.com/question/3776858

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number

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