Question

Solve for d.

Answer 1

EVERYONE STOP NOW

Answer 2

qudrex, try solving this one here, and we will guide you through the process.

Answer 3

okay

Answer 4

So, what is the first step?

Answer 5

no idea. :/ i fail hard at math.

Answer 6

You are trying to solve for \(d\) in\[\sqrt{d^2-11}=5.\]It would be a good idea to start by squaring both sides, do you agree?

Answer 7

that would give me -22 and 10 right?

Answer 8

Howcome? If you square both sides, you will get\[d^2-11=25.\]Do you see why?

Answer 9

no. :/

Answer 10

Let us go back one step: If we square\[\sqrt{d^2-11}=5,\]then we are doing this:\[\left(\sqrt{d^2-11}\right)^2=5^2,\]right?

Answer 11

yeah.

Answer 12

Do you know what\[\left(\sqrt{d^2-11}\right)^2\]is?

Answer 13

no.

Answer 14

The square root sign and the square cancel each other out to give you\[d^2-11.\]Do you know what \(5^2\) is?

Answer 15

10 right?

Answer 16

By definition, \(5^2=5\cdot5=?\)

Answer 17

so 25.

Answer 18

That is why you get\[d^2-11=25.\]Can you solve that for \(d\)?

Answer 19

d = 36? :/

Answer 20

You are close!\[d^2=36.\]We are not done yet, however: we need \(d\) not \(d^2\). Do you know how to fix that?

Answer 21

divide 36 by 2?

Answer 22

No. Just like before, the way to cancel out a square is by taking the square root. That is,\[\sqrt{d^2}=d.\]What you want to do here is\[\sqrt{d^2}=\sqrt{36}.\]What is that?

Answer 23

these are the multipule choice.
4, –4
5, –5
5, 6
6, –6

Answer 24

Oh, well...\[\sqrt{36}=\pm6.\]

Answer 25

so its 6, -6?

Answer 26

Yes.

Answer 27

okay, thankk yoou.

Answer 28

I hope you learned something, though.