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For what real values of $c$ is $x^2 + 16x + c$ the square of a binomial? if you find more than one, then list your values separated by commas.

Respuesta :

danielmaduroh
The Square of a Binomial is always a trinomial. In general, you have the following relationship for a square binomial:

[tex](a+b)^2=a^2+2ab+b^2[/tex]

The trinomial of our problem is:

[tex]x^2+16x+c[/tex]

So we need to find the real value of [tex]c[/tex] such that this is a square of a binomial. Matching the general form with the trinomial of our problem we have the following equations:

[tex]a=x \\ 2ab=16x = 2(x)(8) \\ where \ b=8 \\ \\b^2=c \therefore c=8^2 \therefore \boxed{c=64}[/tex]

Therefore the square of our binomial is:

[tex](x+8)^2=x^2+16x+64[/tex]
MindTrickery

Answer:

-10,10

Step-by-step explanation:

Comparing x^2 + ax + 25 to (x+b)^2 = x^2 + 2bx + b^2, we see that we must have b^2 = 25. This means that b = -5 or b=5.

If b=5, we have (x+5)^2 = x^2 +10x + 25, and if b=-5, then we have (x-5)^2 = x^2 -10x + 25. Therefore, the two values of a for which x^2 + ax + 25$ is the square of a binomial are

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