How do you solve √2x+5−√x−2=3?

1 Answer
Babette
Jan 14, 2016

x=2 and x=38

Explanation:

Square both sides

(√2x+5−√x−2)(√2x+5−√x−2)=(3)2

Expanding the left side we will have

(2x+5)−2√x−2√2x+5+(x−2)=9

Rewrite the product of the radicals on the left hand side

2x+5−2√(x−2)(2x+5)+x−2=9

Combine like terms and expand under the radical on the left

3x+3−2√2x2+x−10=9

Subtract 9 from both sides and add 2√2x2+x−10 to both sides

3x−6=2√2x2+x−10

Now square both sides

(3x−6)(3x−6)=(2√2x2+x−10)2

Expand

9x2−36x+36=4(2x2+x−10)

Distribute the 4 on the right side

9x2−36x+36=8x2+4x−40

Subtract 8x2 from both sides

Subtract 4x from both sides

Add 40 to both sides

x2−40x+76=0

This will factor to

(x−2)(x−38)=0

So x−2=0

x=2

or x−38=0

x=38

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