Squaring both sides twice

[ahspadafora]

I am pulling my hair out over this problem

(√2x+2) - (√x-3)=2

(2x + 2 is all under one radical, same with x-3)


okay so I know you have to isolate one of the radicals so I will add √x-3 to both sides

√2x+2 - √x-3=2
+√x-3= +√x-3
√2x+2= 2 + √x-3

then square each side
(√2x+2)² = (2 + √x-3)²
2x+2 = 2² + 2(2√x-3) + x-3

simplify & combine like terms

2x+2 = 4 + 4√x-3 + x-3
2x+2 = 1 + x + 4√x-3

then isolate the radical again

2x+2 = 1+x + 4√x-3
-1 -1

2x+1 = x + 4√x-3
-x -x

x +1 = 4√x-3

then divide by 4

(x+1)/4 = (4√x-3)/4

x+1 =√x-3
----
4

then I am pretty much lost


please help

 

[DrPhil]

I am pulling my hair out over this problem

(√2x+2) - (√x-3)=2

(2x + 2 is all under one radical, same with x-3)

[That would be clearer if you wrote √(2x+2) - √(x-3) = 2]

okay so I know you have to isolate one of the radicals so I will add √x-3 to both sides

√2x+2 - √x-3=2
+√x-3= +√x-3
√(2x+2) = 2 + √(x-3)

then square each side
(√2x+2)² = (2 + √x-3)²
2x+2 = 2² + 2(2)√(x-3) + x-3

simplify & combine like terms

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

then isolate the radical again

2x+2 = 1+x + 4√(x-3)
-1 .......-1

2x+1 = x + 4√(x-3)
-x ......-x

x +1 = 4√x-3

then divide by 4

(x+1)/4 = 4√(x-3)/4

x+1 =√(x-3)
----
4

You are doing great! Since you have the radical isolated, square both sides!!

 

[ahspadafora]

x+1 =√(x-3)
----
4

okay so from here square both sides

(x+1)^2
______ = √(x-3)²
4²


which becomes

x²+2x+1
______ = x-3
16


then multiply each side by 16 and you'll get

x²+2x+1 =16x-48

add 48 to each side

x²+2x+49=16x

subtract 16x

x²-14x+49=0

when factored becomes

(x-7)(x-7)=0
or x-7=0

x=7


EUREKA!
Thanks Dr.Phil your input helped turn the bulb on in my head

I must of got intimidated by the squaring a fraction =X

 

[lookagain]

x²-14x+49=0

when factored becomes

(x-7)(x-7)=0
or x-7=0

x=7 \(\displaystyle \ \ \ \ \ \) You're not finished! Check this candidate solution by substituting it into the original equation.


EUREKA!
Thanks Dr.Phil your input helped turn the bulb on in my head

I must of got intimidated by the squaring a fraction =X \(\displaystyle \ \ \ \ \) No, all of that is unnecessary, and you're wasting time, energy, and steps.

Here are some of your steps edited together that you needn't have done:

x + 1 = 4√(x - 3)

then divide by 4

(x + 1)/4 = (4√(x - 3))/4


x + 1 = √(x - 3)
-----
4

okay so from here square both sides

(x+1)^2
______ = √(x - 3)²
4²


which becomes

x² + 2x + 1
_________ = x - 3
16


then multiply each side by 16 and you'll get

x² + 2x + 1 = 16x - 48

--------------------------------------------------------------------------------------------------------------
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Recommendations: Don't divide by 4. Then you won't be squaring a fraction. Then you won't be multiplying
each side by a denominator to eliminate the fraction.

Instead:

\(\displaystyle x + 1 \ = \ 4\sqrt{x - 3}\)

\(\displaystyle (x + 1)^2 \ = \ [4(\sqrt{x - 3} \ )]^2\)

\(\displaystyle (x + 1)(x + 1) \ = \ (4)^2(x - 3)\)

\(\displaystyle x^2 + 2x + 1 \ = \ 16(x - 3)\)

\(\displaystyle x^2 + 2x + 1 \ = \ 16x - 48 \)

Continue . . .



You avoid forming a fraction, squaring the fraction, undoing the fraction by multiplying each side by the denominator, etc.