Question

Question in reply

Answer 1

Solve for x.
\[\sqrt{2x + 3} = \sqrt{6x - 1}\]


1
2
3
\[\sqrt{6}\]

Answer 2

can you show me how to get the answer step by step?

Answer 3

Sure. I'm a little confused, though - what are the 1 2 3 and the squr. root supposed to mean?

Answer 4

*Sorry - the squr. root of six

Answer 5

those are the ones to pick from.

A. 1
B. 2
C. 3
D. \[\sqrt{6}\]

Answer 6

OH, okay. My mistake :)

Answer 7

lol it's ok

Answer 8

Alright. Let me get my square root thinking cap on. :)

Now, the important thing to remember, here, is to not let the expressions with the variables scare you. Just take a step back, and look at what's going on. Okay, so we can see that both sides of the equation are being square-rooted. Remember that the OPPOSITE of square root is to SQUARE something? Well, that's what we're going to do to both sides. Now, when you square each expression, remember to put them in brackets, like this: (2x+3) and (6x-1). Okay, so let's square each expression. For the first one: (2x+3)^2 --> (2x+3)(2x+3) --> distribute: 4x^2+6x+6x+9 --> simplify: 4x^2+12x+9 Now for the second expression: (6x-1)^2 --> (6x-1)(6x-1) --> distribute: 36x^2+-6x+-6x+1 --> simplify: 36x^2+-12x+1 Okay, now that we've squared each expression, we need to set them back on each side of the equation, and see what we have; 4x^2+12x+9 = 36x^2+-12x+1 Finally, we're ready to isolate the like numbers and variables on sides of the equation, and solve for x. Here we go, step by step:

1. Subtract 1 from right and left side. This leaves: 4x^2+12x+8=36x^2+-12x
2. Add -12x to each side of the equation. This leaves: 4x^2+24x+8=36x^2
3. Subtract 4x^2 from each side. This leaves: 24x = 32x^2
4. Subtract 24x from both sides. This sets the equation equal to 0. This leaves: 32x^2 - 24x = 0
5. Factor out common terms. We can see that an x can be factored out, as well as an 8. We now have: 8x(4x-3)=0

Okay, see if you can do any more, from here. Let me know if you need more help. :)

Answer 9

I'm sorry - typo: step 2, where it says "Add -12x to each side" should be "Add 12x to each side."

Answer 10

Thanks, Hero!

Answer 11

0. Write Original Problem: \[\sqrt{2x+3} = \sqrt{6x-1}\]1. Square Both Sides: \[2x + 3 = 6x - 1\] 2. Subtract 2x from both sides and add 1 to both sides: \[3 + 1 = 6x - 2x\] 3. Perform the indicated operations: \[4 = 4x\] 4. Divide both sides by 4: \[4/4 = x\]\[1 = x\]

Answer 12

By the the way, when you square a root, both the square and it's root cancels because they are inverses of each other.

Answer 13

Aha - I forgot about that! Thanks so much, Hero!

Answer 14

yw =]

Answer 15

I guess I got mixed up with when you just have expressions that are without a square root. Oops! :)