Question

Which are the roots of the quadratic function f(b) = b2 – 75? Select two options. b = 5 StartRoot 3 EndRoot b = Negative 5 StartRoot 5 Endroot b = 3 StartRoot 5 Endroot b = Negative 3 StartRoot 5 EndRoot b = 25 StartRoot 5 EndRoot

Answer 1

Answer:

b = 5 StartRoot 3 EndRoot

b = Negative 5 StartRoot 3 EndRoot

Step-by-step explanation:

we have

$f(b)=b^{2}-75$

Remember that the root of a function is the value of x when the value of the function is equal to zero

In this problem

The roots are the values of b when the function f(b) is equal to zero

so

For f(b)=0

$b^{2}-75=0$

$b^{2}=75$

square root both sides

$b=(+/-)\sqrt{75}$

Simplify

$b=(+/-)5\sqrt{3}$

$b=5\sqrt{3}$  and $b=-5\sqrt{3}$

therefore

b = 5 StartRoot 3 EndRoot

b = Negative 5 StartRoot 3 EndRoot

Answer 2

The roots of the quadratic function are 5√3 and -5√3

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:

$f(b) = b^2 - 75$

The roots of the quadratic function could be calculated when f(b) = 0 :

$0 = b^2 - 75$

$b^2 = 75$

$b = \pm \sqrt{75}$

$b = \pm \sqrt{25 \times 3}$

$b = \pm \sqrt{25} \times \sqrt{3}$

$b = \pm 5 \times \sqrt{3}$

$b = \pm 5\sqrt{3}$

$b = 5\sqrt{3} \texttt{ or } b = -5\sqrt{3}$

$\texttt{ }$

Learn more

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

Answer details

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number