Question

Solve the equation. If necessary, round your answer to the nearest tenth.

4x^2 = 121

A.
2.75, –2.75

B.
5.5, –5.5

C.
11, –11

D.
30.25, –30.25

Answer 1

do you know how to solve something like x^2=9?

Answer 2

It's D

Answer 3

\[x^2=9 \text{ implies } x=\sqrt{9} \text{ or } x=-\sqrt{9} \text{ are solutions }\]you can simplify those solutions

\[x^2=\frac{121}{4} \text{ is what you want \to solve }\]
try taking square roots of both sides like I did
\[x=\frac{\sqrt{121}}{\sqrt{4}} \text{ or } x=-\frac{\sqrt{121}}{\sqrt{4}}\]
do you know the sqrt(121) and the sqrt(4)?

Answer 4

@Cookie_2046 it isn't D

Answer 5

C.

Answer 6

can you tell me what sqrt(121) and sqrt(4) equal?

Answer 7

Yeah.. i don't know that stuff, i'm bad at this

Answer 8

so do you know 11*11=121
and 2*2=4?

Answer 9

Yes

Answer 10

so you should know sqrt(121)=11 and sqrt(4)=2

Answer 11

\[x=\frac{\sqrt{121}}{\sqrt{4}} \text{ or } x=-\frac{\sqrt{121}}{\sqrt{4}}\]
simplify further

Answer 12

the second one

Answer 13

no those are both the solutions

Answer 14

Like i said i don't know this stuff what do i simplify?

Answer 15

I was asking you to simplify both using that sqrt(121)=11 and sqrt(4)=2

Answer 16

so 121 x 4?

Answer 17

do you see the division bar (fraction bar) between the sqrt(121) and the sqrt(4)

Answer 18

that doesn't mean multiply

Answer 19

and also what happen to the sqrt( ) part

Answer 20

for example
\[x^2=\frac{9}{4} \text{ \implies } x=\frac{\sqrt{9}}{\sqrt{4}} \text{ or } x=-\frac{\sqrt{9}}{\sqrt{4}} \\ \text{ simplifying both of these solutions we get } x=\frac{3}{2} \text{ or } x=- \frac{3}{2}\]
see all I did was replace sqrt(9) with 3
and sqrt(4) with 2