Question

Can someone help me with some Algebra 2 questions?

Answer 1

\[\sqrt{2x-3}=5\]

Answer 2

Solve for x.

Answer 3

the square root of what number is 5?

Answer 4

2?

Answer 5

sqrt(2) = 5 ?? no

Answer 6

Well, I don't know. I'm stupid..seriously when it comes to algebra

Answer 7

this isnt algebra .... there is an algebraic want to approach it, but it would also be useful to know what a square and square root tell us

Answer 8

think about what the question is asking for:

the square root of some number is 5, therefore some number is equal to 5^2

Answer 9

if we square both sides, we can unwrap the square root
\[(\sqrt{...})=k\]
\[(\sqrt{...})^2=k^2\]
\[(...)=k^2\]

Answer 10

i don't know.............

Answer 11

i'm so frustrated right now it's not even funny

Answer 12

5^2 is notation that tells us how many times to multiply a number with itself ....

5^2 means; 5*5

Answer 13

if \(5^2 = n\), then \(\sqrt n = 5\)

Answer 14

this is all foreign language..... you don't have to help me if you don't want to i'm obviously an idiot.

Answer 15

lets assume for the moment that 5^2 = 25

Answer 16

then the sqrt(25) = 5

they give us; sqrt(2x-3) = 5

this tells us that 2x-3 has to equal 25

2x-3 = 25, should be readily solvable for x

Answer 17

\[\sqrt{2x-3}=5~:~square ~both~sides\]
\[(\sqrt{2x-3})^2=5^2\]
\[2x-3=25\]
that should be the hardest part about this ... i trust you can work the rest of it out

Answer 18

No i can't like i said i'm terrible at this and i'm probably going to fail, but thanks anyway

Answer 19

@KelseyKaup123 , amistre did all the hard work, not it is just simple Algebra or Pre-Algebra even! You can do it!!
you want to get "x" all by itself ... so you need to shift things around to do that

first take that "-3" and switch it to the other by add it +3 to both sides, which will then give you 2x = 25+3

Can you solve for that?