Solving Radical expressions

**roachbeats** · February 18th 2009, 07:58 PM

I think that I did this one correctly as well but I could not figure out the factoring.
√x+2 + √3x+7=1
√3x+7 = 1-√x+2
√(3x+7)^2 =(1-√x+2)^2
3x+7=1-x = 2-2√x+2
3x+7 = -x+3-2√x+2
2x +4 = (-2√x+2)
(2x+4)^2= (2√x+2)^2
4x^2+16x+16=4(x+2)
4x^2+12x-8=0

**earboth** · February 18th 2009, 09:35 PM

Originally Posted by roachbeats

I think that I did this one correctly as well but I could not figure out the factoring.
√x+2 + √3x+7=1
√3x+7 = 1-√x+2
√(3x+7)^2 =(1-√x+2)^2
3x+7=1-x = 2-2√x+2
...

$\left(\sqrt{3x+7}\right)^2=\left(1-\sqrt{x+2}\right)^2$

$3x+7=1-2\sqrt{x+2} + x+2$

$\left(2x+4\right)^2 = \left(\sqrt{x+2}\right)^2$

$4x^2+16x+16 = x+2$

$4x^2+15x+14=0$

$x=\dfrac{-15\pm\sqrt{(-15)^2-4\cdot 4 \cdot 14}}{2\cdot 4}$

$x=-2~\vee~x = -\frac74$

Now plug in these solutions into the original equation and check which is a correct solution:

$x = -2~\implies~\sqrt{-2+2} + \sqrt{3\cdot(-2)+7} = 1$ Therefore $\boxed{x = -2}$ is a solution.

$x = -\frac74~\implies~\sqrt{-\frac74+2} + \sqrt{3\cdot \left(-\frac74 \right)+7} = \frac12 + \sqrt{\frac74}\neq 1$ Therefore $x = -\frac74$ is no solution.

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