How do you solve $\sqrt{2 x + 3} - \sqrt{x + 1} = 1$ $sqrt(2x+3)-sqrt( x+1) =1$ ?

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Answer 1 · 2018-04-29T19:26:16

The two solutions are $x = 3$ $x=3$ and $x = - 1$ $x=-1$ .

Isolate one of the radicals, square both sides, isolate the other radical, then square both sides again:

$\sqrt{2 x + 3} - \sqrt{x + 1} = 1$ $sqrt(2x+3)-sqrt(x+1)=1$

$\sqrt{2 x + 3} = 1 + \sqrt{x + 1}$ $sqrt(2x+3)=1+sqrt(x+1)$

${(\sqrt{2 x + 3})}^{2} = {(1 + \sqrt{x + 1})}^{2}$ $(sqrt(2x+3))^2=(1+sqrt(x+1))^2$

$2 x + 3 = 1 + 2 \sqrt{x + 1} + x + 1$ $2x+3=1+2sqrt(x+1)+x+1$

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

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

$\frac{{(x + 1)}^{2}}{2^{2}} = {(\sqrt{x + 1})}^{2}$ $(x+1)^2/2^2=(sqrt(x+1))^2$

$\frac{x^{2} + 2 x + 1}{4} = x + 1$ $(x^2+2x+1)/4=x+1$

$x^{2} + 2 x + 1 = 4 x + 4$ $x^2+2x+1=4x+4$

$x^{2} - 2 x - 3 = 0$ $x^2-2x-3=0$

$(x - 3) (x + 1) = 0$ $(x-3)(x+1)=0$

$x = 3, - 1$ $x=3,-1$

Those are the solutions to the problem. Hope this helped!

How do you solve sqrt(2x+3)-sqrt( x+1) =1√2x+3−√x+1=1sqrt(2x+3)-sqrt( x+1) =1?