Question

32x - 4 = 4x2 + 60 For the equation shown, choose the description of the solutions.

Answer 1

Answer:

Two real and equal solutions.

Step-by-step explanation:

We have been given an equation $32x-4=4x^2+60$. We are asked to find the type of solutions of our given equation.

First of all, we will gather all terms on left side of our equation using opposite operations:

$-4x^2+32x-4-60=4x^2-4x^2+60-60$

$-4x^2+32x-64=0$

Dividing our equation by -4 we will get,

$x^2-8x+16=0$

To determine the type of solutions for our given problem, we will use discriminant formula.

$D=b^2-4ac$

b = Coefficient of x term,

a = Leading coefficient,

c = Constant term.

Upon substituting our given values in discriminant formula, we will get,

$D=(-8)^2-4(1)(16)$

$D=64-64$

$D=0$

The interpretations for the value of D are as follows:

$D=0;\text{ Two real and equal roots}$

$D>0;\text{ Two real and distinct roots}$

$D$

Since the value of D is equal to 0 for our given equation, therefore, our given equation will have two real and equal roots or solutions.

Answer 2

All to one side:

4x^2-32x+64 = 0,

divide by 4:

x^2 -8x + 16 = 0

apply formula or see that it is (x-4)^2!, so only x =4 is a root and it is double.