Radical equations? Pleeease help?

Question

Can anyone help me with these radical equations?
[Radical]4x+17 + [Radical] x +7 = [Radical] x + 2

Where [Radical] means the radical, and 4x+7 and x+ 7 are under separate radicals
Please!

Answer 1

Solve for x.

√(4x+17) + √(x +7) = √(x + 2)

Square both sides.
(4x + 17) + 2√{(4x+17)(x + 7)} + (x + 7) = x + 2
(5x + 24) + 2√{(4x+17)(x + 7)} = x + 2

Isolate the square root.
2√{(4x+17)(x + 7)} = -4x - 22
2√(4x² +45x + 119) = -4x - 22
√(4x² +45x + 119) = -2x - 11

Square both sides.
4x² +45x + 119 = 4x² + 44x + 121
x = 2

However, this solution is invalid. It is an extraneous solution introduced by the two times both sides were squared. There is no real solution. Interestingly though, if the equation subtracted the second radical from the first, the solution x = 2 would be valid.

√(4x+17) - √(x +7) = √(x + 2)
√(4*2 +17) - √(2 +7) = √(2 + 2)
√25 - √9 = √4
5 - 3 = 2

I can't help but wonder if your problem was supposed to have the minus sign.

Answer 2

Assuming that x+2 on the right is all under a radical sign...

sqrt(4x+17) + sqrt(x+7) = sqrt(x+2)

Square both sides of the equation. Be sure to expand the left side correctly:

(sqrt(4x+17))^2 + 2 * sqrt(4x+17)sqrt(x+7) + (sqrt(x+7))^2 = (sqrt(x+2))^2

Simplify:

(4x+17) + 2*sqrt((4x+17)(x+7)) + (x+7) = x+2
5x + 24 + 2sqrt(4x^2+45x+119) = x+2

Isolate the radical term:

2sqrt(4x^2+45x+119) = -4x-22
sqrt(4x^2+45x+119) = -2x-11

Square it again:

4x^2+45x+119 = 4x^2+44x+121
x=2

Check in the original equation:

sqrt(4*(2)+17) + sqrt(2+7) = sqrt(2+2)
sqrt(25)+sqrt(9)=sqrt(4)
5+3=2

Apparently I didn't check my original answer very well. I added -2 instead of subtracting it at one point, and the numbers came out completely differently. Nevertheless, when you check the solution of x=2, you get a nonsensical result, which is a product of all that squaring. Therefore, your equation has no solution.