Question

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

Answer 1

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

Answer 2

The zeros of the function f(x) are :

x = ( -5 ± √5 ) / 2

Further explanation

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

D = b² - 4 a c

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!

Given:

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

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

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

$0 = x^2 + 5x + 5$

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

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

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

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

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

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

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

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

$\texttt{ }$

Learn more

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

Answer details

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number