Question

Solve the equation.
Square root of 5x-1 = Square root of 4x+9

Answer 1

\[\sqrt{5x-1}=\sqrt{4x+9}\]

Answer 2

Start by doing the inverse, square both sides to cancel out the square root.

\(\ (\sqrt{5x - 1})^2 = (\sqrt{4x + 9})^2\)

5x - 1 = 4x + 9
+1 +1

5x = 4x + 10
-4x -4x

x = 10


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\(\\ \sqrt{5(10) - 1} = \sqrt{50 - 1} = \sqrt{49} = 7 \)

\(\ \sqrt{4(10) + 9} = \sqrt{40 + 9} = \sqrt{49} = 7 \)

They both equal 7 when you check if 10 is correct. Therefore x does indeed equal 10 \( \color{green}\checkmark\)

Answer 3

Thanks!
Solve the equation for x. Identify any extraneous solutions
\[y=\sqrt{-6y}\]

Answer 4

@tHe_FiZiCx99

Answer 5

x? There's only a y

Answer 6

Sorry Solve the equation for y. Identify any extraneous solutions

Answer 7

@Hero

Answer 8

Are you sure it's y = √-6y ?

Answer 9

positive the answer choices are stuff like
1.0 and -6 are solutions to the original equation

Answer 10

Really?

It's a constant back and fourth loop,

y = \(\ \sqrt{-6y} \)

square both sides and divide by -6, you would just get;

y^2/-6 = y

And if you solved for the left side you would end up with the original equation.

If anything y = 0


0 = √-6(0)

0 = √0 = 0 .-.

Answer 11

Waait, I \(\ think\) its -6 ♦o♦

Answer 12

@tHe_FiZiCx99, how come you didn't square both sides like you did with the first problem?

Answer 13

What do you mean, I did?

Answer 14

lol, you're right

Answer 15

You did something wrong though

Answer 16

I'm brain storming these in my head, and writing them along. I might have some typos.

Answer 17

You should have ended up with
y^2 + 6y = 0
y(y + 6) = 0

y = 0
y = -6

Answer 18

yeah, typo >_<

Answer 19

I actually did it another way in my head :/

Answer 20

Yeah, y = √6 * √-y = y^2 = 6 * (-y)

y^2 = 6*(-y)
/-y /-y

-y = 6
/-1 /-1

y = -6
y = 0

Yeah somehow I managed to do this in my head and forget it for a bit :<

Answer 21

The only problem is, you're not allowed to divide by the variable you are solving for. The reason is because it actually eliminates one of the solutions.

Answer 22

Yeah, I thought about the zero's but technically I assumed there's an understood 0

Answer 23

Fun problem nonetheless :>

Answer 24

I understand what you're trying to say

If y = 0 then

\(0 = \sqrt{-6(0)}\)
0 = 0

However, specificially with regard to solving for y, there's a proper solving methodology and mathematicians have agreed that dividing by the variable you're solving for is not proper for two reasons:

1. You eliminate a solution
2. y = 0 and dividing by zero is frowned upon.

Answer 25

There's an understood zero, but that doesn't justify dividing by zero

Answer 26

Oh that part, yeah :/ you can't divide by 0's

Answer 27

Thanks for clarifying that though :P

Answer 28

yw

Answer 29

...............mind blown haha anyways i dont get what the answer would be?

Answer 30

y=0 & y=-6

Answer 31

Actually, y = 0 may be the only correct solution.

Answer 32

If we check with y = -6 we get the following:

\(-6 = \sqrt{-6(-6)}\)
\(-6 = \sqrt{36}\)
\(-6 = 6\) FALSE

Answer 33

So y = 0 is therefore the ONLY correct solution.

Answer 34

"Sorry Solve the equation for y. Identify any extraneous solutions"

y = -6 Would be considered an \(\ extraneous \) solution

Answer 35

alright thank you so much could you help with just one more please?

Answer 36

Sure, I'll try

Answer 37

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

Answer 38

Solve the equation for x. Identify any extraneous solutions

Answer 39

o_o that's it?

Answer 40

lol, really?

Answer 41

I think you missed something >.>

Answer 42

Maybe he means \(\sqrt{x} = \sqrt{-3}\)

Answer 43

In that case, you start by squaring both sides.

Answer 44

You might run into a problem while checking though

Answer 45

sorry \[\sqrt{x}=-3\]

Answer 46

Either way, still square both sides and run into a problem while checking

Answer 47

Square both sides, x = 9 >.>

√9 = 3 not -3, that's extraneous

Answer 48

I'm curious ot know the +/- sign with alt keys :3

Answer 49

to*

Answer 50

so their is no real solution

Answer 51

Yeah, there's no solution, "That's extraneous"