How do you simplify $\sqrt{2 x + 2} - \sqrt{x + 2} = 1$ $sqrt(2x+2) - sqrt(x+2) = 1$ ?

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Answer 1 · 2016-07-06T15:43:29

$x = 7$ $x = 7$

Squaring both sides

$2 x + 2 - 2 \sqrt{(2 x + 2) (x + 2)} + x + 2 = 1$ $2x+2-2sqrt((2x+2)(x+2))+x+2=1$

grouping

$3 x + 3 = 2 \sqrt{(2 x + 2) (x + 2)}$ $3x+3=2sqrt((2x+2)(x+2))$

squaring both sides again

$9 (x^{2} + 2 x + 1) = 4 (2 x + 2) (x + 2)$ $9(x^2+2x+1)=4(2x+2)(x+2)$

or

$(x + 1) (x - 7) = 0$ $(x+1)(x-7)=0$

Testing the roots in

$\sqrt{2 x + 2} - \sqrt{x + 2} = 1$ $sqrt(2x+2) - sqrt(x+2) = 1$

we choose $x = 7$ $x = 7$

How do you simplify sqrt(2x+2) - sqrt(x+2) = 1√2x+2−√x+2=1sqrt(2x+2) - sqrt(x+2) = 1?