Question

For which of the following equations are x = 5 and x = –5 both solutions?

Answer 1

Short Answer B
Argument
A
A will give you x = +/- 5i 
x^2 + 25 = 0
x^2 = - 25       Take the square root.
sqrt(x^2) = +/- sqrt(-25)
x = +/- (5)i which is a complex number.


B
Is the answer
x^2 = 25
sqrt(x)^2 = sqrt(25)
x = +/- 5

C
Can't be factored just by looking at it. You can show that C is not true just by putting 5 into the equation
f(x) = x^2 + 10x - 25
f(5) = 25 + 10*5 - 25
f(5) = 50
C is not true.

D
D can be eliminated as C was
f(x) = x^2 - 5x - 25
f(5) = -25 ( l'll let you show this is not true). 5 is not a solution because it does not make f(x) = 0

Answer 2

x² – 25 = 0 has both x = 5 and x = - 5 as its solutions.

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

Let us now tackle the problem!

If a quadratic equation has solution x₁ and x₂ , then we could write the equation as following :

$\large {\boxed {a (x - x_1)(x - x_2) = 0} }$

If x₁ = 5 and x₂ = - 5 , then :

$a (x - 5)(x - (-5)) = 0$

$a (x - 5)(x + 5) = 0$

$a (x^2 - 5x + 5x - 25) = 0$

$a (x^2 - 25) = 0$

If a = 1 , then we get :

$a (x^2 - 25) = 0$

$1 (x^2 - 25) = 0$

$\large {\boxed {x^2 - 25 = 0} }$

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