Question

sqrt 2x+13-5=x

Answer 1

what terms are under the square root?

Answer 2

\(\huge \sqrt{2x+13} - 5 = x\)

Answer 3

What do we typically do when solving for x in regular equations, such as \(2x + 5 = 11\) as an example.

Answer 4

Correct and that's the first step. Use that similar approach and do it with the equation in your problem.

Answer 5

When a term is under a radical, you can't do anything with it just yet. But you are correct in adding 5.

Answer 6

so you would have:
\(\huge \sqrt{2x+13} = x + 5\)

Answer 7

To take out the 2x + 13 from the square root, we need to do the inverse of finding the square root, which is...?

Answer 8

\(2^2 = 4\) \(\leftarrow\) we squared 2 to get to 4
inversely...
\(\sqrt{4} = 2\) \(\leftarrow\) if we find the square root of 4, we get 2 (which we originally squared to get to 4)

Answer 9

so what's the inverse of finding the square root?

Answer 10

yes. so you need to square both sides of the equation
\(\huge (\sqrt{2x + 13})^2 = (x + 5)^2\)

Answer 11

Just a final thought...
most likely you will have a quadratic to factor.

Answer 12

and if you have \((x-5)(x-3)\) as an example, your x values will be \(5\) and \(3\) because a quadratic equation will always equal to 0.

Answer 13

how far did you get?

Answer 14

2x+13=x+5

Answer 15

I hope you mean
\[ 2x+13= (x+5)^2 \]

Answer 16

ye

Answer 17

do you know how to multiply out
(x+5)(x+5)
people use FOIL (if you learned that)

Answer 18

It's a bit complicated to explain here. But if you have time
https://www.khanacademy.org/math/algebra-basics/quadratics-polynomials-topic/multiplying-binomials-core-algebra/v/square-a-binomial
will show you how

Answer 19

FOIL: \[(a + b)(a + b) = a(a) + a(b) + b(a) + b(b) = \bf{a^2 + ab + ba + b^2}\]