Answer:
Two real and equal solutions.
Step-by-step explanation:
We have been given an equation [tex]32x-4=4x^2+60[/tex]. 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:
[tex]-4x^2+32x-4-60=4x^2-4x^2+60-60[/tex]
[tex]-4x^2+32x-64=0[/tex]
Dividing our equation by -4 we will get,
[tex]x^2-8x+16=0[/tex]
To determine the type of solutions for our given problem, we will use discriminant formula.
[tex]D=b^2-4ac[/tex]
b = Coefficient of x term,
a = Leading coefficient,
c = Constant term.
Upon substituting our given values in discriminant formula, we will get,
[tex]D=(-8)^2-4(1)(16)[/tex]
[tex]D=64-64[/tex]
[tex]D=0[/tex]
The interpretations for the value of D are as follows:
[tex]D=0;\text{ Two real and equal roots}[/tex]
[tex]D>0;\text{ Two real and distinct roots}[/tex]
[tex]D<0;\text{ No real roots}[/tex]
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.
