Question

Can someone help me solve 2x^2=200 using the square root property?

Answer 1

first you have to solve for x

Answer 2

2 is multiplying by x^2
to cancel that you have to do opposite of multiply

Answer 3

Its inbetween 14.1 and 14.2

Answer 4

imagine you replaced the \(x^2\) with \(y\). This would leave you with:\[2y=200\]Can you solve this to find \(y\)?

Answer 5

Yes. y would equal 100, yes?

Answer 6

correct. so now you are left with:\[y=x^2=100\]so you can find \(x\) by solving:\[x^2=100\]

Answer 7

Okay so x=10. Thank you.

Answer 8

perfect! :)

Answer 9

There's something missing

Answer 10

?

Answer 11

do you mean plus/minus

Answer 12

There are 2 solutions. Yes

Answer 13

normally at this level the square root is taken to be the positive square root

Answer 14

for example, x^2 = 25 has two solutions since both x = 5 and x = -5 make it true

Answer 15

@CelesteSelene13 - have you been taught about square roots having both a positive and negative value yet?

Answer 16

\(\Large x^2 = 25\) leads to \(\Large x = \pm \sqrt{25}\)

normally the square root only produces one output, so that's why the plus/minus is needed

Answer 17

@CelesteSelene13 - if you have been taught about positive and negative values for square roots then take Jim's result. Otherwise just stick with the positive value.

Answer 18

@CelesteSelene13 - do you understand what Jim was trying to say?

Answer 19

@asnaseer Yes, yes I have and yes, I do understand. Thank you both

Answer 20

OK - in that case I apologise @jim_thompson5910 - you were indeed correct here :)