If $a$ is a constant such that $4x^2 + 14x + a$ is the square of a binomial, then what is $a$?

Question

Please help ASAP. Really need it. Thank you!

Answer 1

4x²+14x+a=4(x²+7x/2+a/4)=

4(x+7x/2+49/16-49/16+a/4)=

4((x+7/4)²+a/4-49/16).

So if the binomial is 2(x+7/4)=2x+7/2, then a/4=49/16 and a=49/4.

Quick check: 4x²+14x+49/4=4(x²+7x/2+49/16)=(2(x+7/4))²=(2x+7/2)².

Merry Christmas!

Answer 2

49/4

Answer 3

??? I don't know the answer?

Answer 4

We know that (a+b)^2 is a square of a binomial, which equals (a^2+2ab+b^2).

a^2=4, so a=2. So, we have (2x+b)^2=(4x+4bx+b^2).

So, 4b=14. Dividing each side by 4, we get 3.5.

Now we have (2x+3.5)^2.

In this problem, b^2 = a. So, 3.5^2=12.25.

a=12.25.