what is the best and faster way in answering radical equation?

Question

for example:
x = √x+7 + 5
square root of x+7

for example:
x = √x+7 + 5
square root of x+7

isn't it better to do it this way

transpose 5 into the other side

x-5=√x+7
then square both sides
x2-25=x+7
x square
then transpose the other x to the other side
then transpose -25 to the other side
so the answer should be
x=32
right?

wrong answer sorry it was supposed to
x= -17

sorry

x=18

sorry

x=18

sorry

x=18

Answer 1

x = √x + 7 + 5

x - 5 = √x + 7 + 5 - 5

x - 5 = √x + 7

(x - 5)² = √x + 7

x² - 10x + 25 = x + 7

x² - 10x + 25 - x = x + 7 - x

x² - 11x + 25 = 7

x² - 11x + 25 - 7 = 7 - 7

x² - 11x + 18 = 0

(x - 9)(x - 2)

- - - - - - -

roots

x - 9

x - 2

- - - - - -s-

Answer 2

I'm not sure this is the fastest, but I'd do this:
1) subtract 5 from both sides, so that the equation is

x -5 = sqrt(x+7)

The reason for doing this is that I want the radical all by itself for the next step:

2) square both sides:

(x-5)^2 = x+7

3) expand all the terms:

x^2 + 25 - 10x = x+7

4) put everything on the left:
x^2 - 11x + 18 = 0

5) use the quadratic formula to solve for x:
(x-9)*(x-2) = 0
so x = 9, or x = 2 is a possible solution

6) However, there was really ambiguity in the square root, which could have been a plus or minue, so we have to test both answers:
9 = sqrt(9+7) + 5? 9 = sqrt(16) + 5 (yes)
and
2 = sqrt(9) + 5 (no)
So the answer is x = 9

I guess this is not very fast...

Answer 3

Your own solution makes one of those classic mistakes in squaring a binomial : (x-5) squared DOES NOT equal
x^2 + 25; nor does it equal x^2 - 25. Please evade this potential bad habit!

George's answer-approach gives the POTENTIAL answers; however, checking shows that 2 is an invalid
solution. (Notice he does sq the binomial correctly, tho.)

It's assumed that your original problem has the quantity
(x+7) all under the radical sign.

Your strategy is to isolate that term (with the variable/unknown under the radical sign) on one equation
side; then raise each side of the equation to the power
of the index for the radical. (In your case the index is
understood to be 2.) That process clears the radical sign from your equation. I think you feel ok from there (?).

Answer 4

Look for a way to get rid of the surd.
x-5 = sqr (x+7)
(x-5)^2 = x +7
x^2 - 10x + 25 = x +7
x^2 - 11x + 18 = 0
(x-2)(x-9)=0
x=2 or 9

Answer 5

Before you square the equation, check that both member are either positive or negative, so that you dont introduce roots

x-5 = sqrt (x+7)

The square root is positive or 0 (and x must be at least -7), so, x >=5.

Ana