Question

How do I solve, X^2 - 25?

Answer 1

By factoring

Answer 2

(x+5)(x-5)
x = -5
x = 5

Answer 3

How did you get that?

Answer 4

hmm was there an equal there? what are you supposed to do with that anyway?

Answer 5

\(\bf x^2-25\) will have no solution, because it's just a bare polynomial

Answer 6

@jdoe0001 You have to make it into a binomial pair, but I don't know how to do that..

Answer 7

ahhh then @shamil98 is correct, you're asked to FACTOR it then

Answer 8

well... with the exception of the = part =)

Answer 9

I have no idea how he/she got that though... :'(

Answer 10

(x+5)(x-5) = 0
x+5 = 0
x = -5
x -5 = 0
x = 5

Answer 11

\(\bf x^2-25\qquad \textit{recall that}\quad 25 = 5^2\qquad thus\\ \quad \\
x^2-25\implies x^2-5^2\qquad \textit{keep in mind that }\quad a^2-b^2 = (a-b)(a+b)\\ \quad \\
x^2-5^2\implies (x-5)(x+5)\)

Answer 12

Oh ok thanks I get it now.

Answer 13

How would I solve 64W^2 - 49?

Answer 14

what is the square root of 64 and 49?

Answer 15

\(\bf 64w^2 - 49\qquad \textit{notice}\quad 64 = 8^2\qquad 49 = 7^2\\ \quad \\
64w^2 - 49\implies 8^2w^2-7^2\implies (8w)^2-7^2\\\quad \\\qquad \textit{then use }\quad a^2-b^2 = (a-b)(a+b)
\)

Answer 16

So is it (8w - 7)(8w +7) And do I just use the square root on the problems like this?

Answer 17

if the coefficient of x^2 is a perfect square and you are subtracting a constant that is a perfect square then yes.

Answer 18

I have this problem 125a^2 - 320 I can't square these numbers what do I do to solve it?

Answer 19

get the gcf

Answer 20

The GCF of 125 & 320?

Answer 21

Its 5 right? and now what am I soupose to do?