Question

ASAP MEDAL!
solve the equation sqrt11x-5=sqrt9x+7

Answer 1

Is this it?

\(\sqrt{11x-5} = \sqrt{9x+7}\)

Have you first considered the Domain?

Answer 2

yes, and no. can you also please explain step by step as to how to get the answer? I would like to understand it.

Answer 3

You must consider the Domain.

\(11x-5 \ge 0\) Do you know why?
\(9x+7 \ge 0\) Do you know why?

Answer 4

i do not. can you explain

Answer 5

Can you find the square root of a negative number?

Answer 6

honestly im so bad at math... im sorry I sound super dumb it;s just not my strong subject..

Answer 7

Never say that again. Now, answer my question.

\(\sqrt{4} = 2\) because 2*2 = 4

What is \(\sqrt{-4}\)? Can you multiply two identical REAL numbers and get -4?

Answer 8

-2*-2

Answer 9

Does that work? (-2)(-2) = +4 We did NOT get -4. Try again.

Answer 10

oh no then? cause -2*2=-4 but they aren't the same?

Answer 11

Perfect!!! If we are going to stay in the Real Numbers, the is NO Square root of nagative numbers. Keep this important fact in mind.

We have \(\sqrt{11x-5}\), so 11x - 5 BETTER NOT BE negative. Agreed?

Answer 12

yes

Answer 13

Then, \(11x - 5 \ge 0\) or \(11x\ge 5\) or \(x \ge 5/11\).

You do the other side. \(\sqrt{9x+7}\)

Answer 14

Im so confused as to how you did that...

Answer 15

What is it that needs not to be negative?

Answer 16

so 9x+7>=0 or 9x>=7 or x>=7/9?

Answer 17

-7/9

Do you see the error?

Answer 18

The second to last thing we need is a little thing called equivalence.

\(If \sqrt{a} = \sqrt{b}\;then\;a = b\). Some might construe this as Squaring Both Sides, but that idea leads to more errors than it should.

We have \(\sqrt{11x-5} = \sqrt{9x+7}\), so 11x-5 = 9x+7. Solve for x and make sure the result is in the established Domain, x >= -7/9