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1 month ago

11

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

Root 5 Endroot b = 3 StartRoot 5 Endroot b = Negative 3 StartRoot 5 EndRoot b = 25 StartRoot 5 EndRoot

2 answers:

zzz [12.3K]1 month ago

8 0

Answer:

b = 5√3

b = -5√3

Step-by-step explanation:

We have

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

Keep in mind, the root of a function corresponds to the value of x at which the function's output is zero.

In this case

The roots of the equation are the b values for which f(b) equals zero.

Thus

For f(b)=0

$b^{2}-75=0$

$b^{2}=75$

Take the square root of both sides

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

Now simplify

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

$b=5\sqrt{3}$ and $b=-5\sqrt{3}$

So we find that

b = 5√3

b = -5√3

Leona [12.6K]1 month ago

6 0

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

Further explanation

To find the Discriminant for a quadratic formula, ( ax² + bx + c = 0 ), use this:

D = b² - 4ac

Based on the Discriminant's value, we can infer the number of solutions the equation may have following these criteria:

D < 0 → No Real Roots

D = 0 → One Real Root

D > 0 → Two Real Roots

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

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

Now let’s solve the problem!

Given:

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

The roots of the function can be determined 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:
Determine the Discriminant:
Formula of Quadratic Equations:

Answer details

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic, Equation, Discriminant, Real, Number

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