Answer:
for all values of x = 2 is the double product of the binomials x+2 and x–2 is 16 less than the sum of their squares
Step-by-step explanation:
By Using the difference of two square:
The product of the binomials:
[tex](x+2)(x-2)=(x^2-4)[/tex]
Double the product yields = [tex]2(x^2-4)[/tex] ----- equation (1)
However, the sum of their squares can be expressed as follows:
[tex](x+2)^2+(x-2)^2[/tex]
16 less than that will be:
[tex](x+2)^2+(x-2)^2-16=0[/tex] ----------- equation (2)
NOW; the question says for what value of the "x" are those two equation equal to each other;
To tackle the problem; we equate equation (1) and equation (2) together.
So; now we can have:
[tex]2(x^2-4)[/tex] = [tex](x+2)^2+(x-2)^2-16[/tex]
[tex]2(x^2-4) = (x^2+2x+2x+4)+(x^2-2x-2x+4)-16[/tex]
[tex]2x^2-8 = (x^2+4x+4)+(x^2-4x+4)-16[/tex]
[tex]2x^2-8 = 2x^2+8-16[/tex]
[tex]2x^2-8 = 2x^2-8[/tex]
In this scenario; we found out that the left-hand side is equal to the right-hand side.
So;
[tex]2x^2-8 = 0\\2x^2=8\\x^2 =\frac{8}{2} \\x^2 = 4\\x =\sqrt{4} \\x=2[/tex]
We therefore conclude that for all values of x = 2 is the double product of the binomials x+2 and x–2 is 16 less than the sum of their squares
Also, we can say that for all values of "x" = [tex]( \infty, + \infty)[/tex] for all intervals or [tex](x\in \mathbb{R})[/tex] for set notations.
