For what value of x is the double product of the binomials x+2 and x–2 is 16 less than the sum of their squares? pls help

Question

For what value of x is the double product of the binomials x+2 and x–2 is 16 less than the sum of their squares?

pls help I’m confused

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Vivian 4 weeks 2021-09-19T04:47:31+00:00 1 Answer 0

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    2021-09-19T04:49:28+00:00

    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:

    (x+2)(x-2)=(x^2-4)

    Double the product yields = 2(x^2-4)    —– equation (1)

    However, the sum of their squares can be expressed as follows:

    (x+2)^2+(x-2)^2

    16 less than that will be:

    (x+2)^2+(x-2)^2-16=0             ———– 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:

    2(x^2-4)   = (x+2)^2+(x-2)^2-16

    2(x^2-4) = (x^2+2x+2x+4)+(x^2-2x-2x+4)-16

    2x^2-8 = (x^2+4x+4)+(x^2-4x+4)-16

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

    2x^2-8 = 2x^2-8

    In this scenario; we found out that the left-hand side is equal to the right-hand side.

    So;

    2x^2-8 = 0\\2x^2=8\\x^2 =\frac{8}{2} \\x^2 = 4\\x =\sqrt{4} \\x=2

    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” = ( \infty, + \infty) for all intervals or (x\in \mathbb{R}) for set notations.

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