X^2+ 8x+ How to make it a perfect square

Hey there! I'm happy to help!

A perfect square means that this equation is the product of two identical binomials (this is a quadratic due to the x^2, so it is made up of two binomials) so for example (x+2)(x+2) would be x^2+4x+4 which is a perfect square as you can square root x^2+4x+4 and (x+2) and not some other crazy thing.

So, we have most of the final quadratic but we need the one constant number on the end. How are we going to find it?

If you are familiar with quadratics, you probably know that adding the two numbers in the binomials gives you b and multiplying them gives you c with a quadratic ax^2+bx+c because of the foil method (if you don't know what that it is, just comment on the answer and I can explain it to you). Basically what that tells us is that our binomial numbers are going to add up to be eight (and we also know that they must be the same number as we want a perfect square). We know that 4+4 is equal to eight, so that means that our binomial pair will look like this:

(x+4)(x+4)

And we also know that multiplying them gives you c, so 4×4=16 and there is our c. We just throw it on at the end of the equation.

x^2+ 8x+16

If you have any more questions, feel free to ask me. Have a great day! :D

answer: sorry

step-by-step explanation: can you tell me pls how to find the rational of two integers

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.