HOW do you solve this equation?

Question

The square root of x+1 plus the square root of 3x equals the square root of 9x-2.

I know you square both sides...but then I got x+1+3x = 9x-2 and it wasn't correct. If anyone can just help me fix that part, that would be great and I can go from there.

To HayHarBr: Well, what if you square two things that are subtracted? Same thing?

airae soares : That answer is incorrect. That's the exact answer I got on my test, and I lost all possible points for that question.

Answer 1

The trouble is that when you square two things that are added, you can't just square each of them. It has to be done by FOIL and my trouble was I missed an x EDIT:

(sqrt x+1 + sqrt 3x)(sqrt x+1 + sqrt 3x)

(x+1) + 2 sqrt(3x(x+1)) + 3x

2 sqrt (3x(x+1)) + 4x + 1= 9x - 2

get the added stuff over to the right and square both sides again

I hope that is all you needed to get you going

Same if subtracted

Look at it with simple numbers
[sqrt (9) - sqrt 4] squared is 3-2 squared or 1

but if you just squared each square root youd get 9 - 4 = 5

(sqrt 9 - sqrt 4)(sqrt 9 - sqrt4) = 9 - 2 sqrt 36 + 4

= 9 - 2(6) + 4

= 13 - 12
= 1

Answer 2

sqrt(x + 1) + sqrt(3x) = sqrt(9x - 2)

Square both sides; remember that you have to FOIL the left hand side. Remember that squaring a square root (as per the right hand side) results in the removal of the square root symbol.

[sqrt(x + 1) + sqrt(3x)]^2 = 9x - 2

FOIL out the left hand side, to get

[sqrt(x + 1)]^2 + 2sqrt(x + 1)sqrt(3x) + [sqrt(3x)]^2 = 9x - 2

Again, by the earlier principal of squares cancelling square roots,

(x + 1) + 2sqrt(x + 1)sqrt(3x) + 3x = 9x - 2

Bring everything but the term with the square roots to the right hand side.

2sqrt(x + 1)sqrt(3x) = 9x - 2 - 3x - x - 1
2sqrt(x + 1)sqrt(3x) = 5x - 3

Note that whenever we have sqrt(a)sqrt(b), we can combine it into a single radical, sqrt(ab). That's what we do here.

2sqrt ( 3x(x + 1) ) = 5x - 3

Multiply out the inside, to get

[2sqrt(3x^2 + 3x)] = 5x - 3

Now, square both sides.

[2sqrt(3x^2 + 3x)]^2 = (5x - 3)^2

4(3x^2 + 3x) = 25x^2 - 30x + 9

12x^2 + 12x = 25x^2 - 30x + 9

Move everything over to the right hand side.

0 = 13x^2 - 42x + 9

Now, factor.

0 = (13x - 3) (x - 3)

Therefore, equating those two binomials to 0,

13x - 3 = 0
x - 3 = 0

This leads to the solution x = {3/13, 3}

BUT WAIT! You can't assume that both of these values work. What you have to do to verify that they work is TEST them into the ORIGINAL EQUATION, sqrt(x + 1) + sqrt(3x) = sqrt(9x - 2).
If we end up taking the square root of a negative number, we discard that result.

For sqrt(x + 1) + sqrt(3x) = sqrt(9x - 2):

Test x = 3/13:
LHS = sqrt(3/13 + 1) + sqrt(3(3/13) = sqrt (16/13) + sqrt(9/13)
= 4sqrt(1/13) + 3sqrt(1/3) = 7sqrt(1/3)
RHS = sqrt(9(1/13) - 2) = sqrt(9/13 - 2) = sqrt (9/13 - 26/13), which is a negative number. Reject x = 3/13.

Test x = 3.

LHS = sqrt (3 + 1) + sqrt(3(3)) = sqrt(4) + sqrt(9) = 2 + 3 = 5
RHS = sqrt(9x - 2) = sqrt(9(3) - 2) = sqrt(25) = 5

Therefore x = 3 is your only answer.

Answer 3

YES, square both sides but remember (a+b)squared = a^2 +b^2 +2ab. So x+1 + 3x +2sqrt((x+1)(3x) = 9x -2

2sqrt(3x^2 + 3x) = 5x - 3 Now, square again.

12x^2 +12x = 25x^2 +9 - 30x
13x^2 - 42x + 9 = 0

Answer 4

Domain: x ≥ 2/9

Square both sides of the original equation,
x+1+3x+2√[(x+1)(3x)] = 9x-2
2√[(x+1)(3x)] = 5x-3, x ≥ 3/5 (adjusted domain)

Square both sides of above equation,
12x^2+12x = 25x^2-30x+9

Simplify,
13x^2-42x+9 = 0

Factor,
(13x-3)(x-3) = 0

x = 3/13 is not a solution because it violates x ≥ 3/5.

x = 3 is the only solution

Answer 5

square root of x+1 plus the square root of 3x equals the square root of 9x-2
[\/(x+1)]² + (\/3x)² = \/(9x-2)²
x+1 +3x = 9x - 2
4x - 9x = -2 -1
-5x = -3
x = 3/5
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Answer 6

√(x+1) + √3x = √(9x - 2)
(x + 1) + 2(√(x+1)√(3x)) + 3x = 9x - 2
4x + 1 + 2(√(4x + 1)) = 9x - 2