Help with Polynomials and Radicals

**yewchung** · November 29th 2012, 03:15 PM

I am confused about the following problem, and cannot figure out how to solve it without using a graphing calculator.

Sqrt(x+5) - Sqrt(2x+3) = -2

How does one solve such a problem?

**zhandele** · November 29th 2012, 04:18 PM

Square both sides. You'll still have a radical on the LHS. Move everything but the radical term to the RHS, then square again. Should work.

**yewchung** · November 29th 2012, 04:39 PM

Ah. Thanks much.

**zhandele** · November 29th 2012, 04:47 PM

What answer did you get?

**Soroban** · November 29th 2012, 06:21 PM

Hello, yewchung!

I tried the problem and got clumsy answers, but they checked out.
(Well, one of them checked . . . the other was extraneous.)

$\text{Solve for }x\!:\;\sqrt{x+5} - \sqrt{2x+3} \:=\: -2$

Isolate a radical: . $\sqrt{2x+3} \:=\:\sqrt{x+5} + 2$

Square both sides: . $\left(\sqrt{2x+3}\right)^2 \;=\;\left(\sqrt{x+5}+2\right)^2$

n . . . . . . . . . . . . . . . . . $2x + 3 \;=\;x+5 + 4\sqrt{x+5} + 4$

Isolate the radical: . . . . . $x - 6 \;=\;4\sqrt{x+5}$

Square both sides: . . . $(x -6)^2 \;=\;\left(4\sqrt{x+5}\right)^2$

n . . . . . . . . . . . $x^2 - 12x + 36 \;=\;16(x+5)$

n . . . . . . . . . . . $x^2 - 12x + 36 \;=\;16x + 80$

n . . . . . . . . . . . $x^2 - 28x - 44 \;=\;0$

Quadratic Formula: . $x\;=\;\frac{28\pm\sqrt{960}}{2} \;=\;\frac{28\pm8\sqrt{15}}{2} \;=\;14 \pm4\sqrt{15}$

The only root is: . $x \;=\;14 + 4\sqrt{15} \;=\;29.491933385\hdots$

**yewchung** · November 29th 2012, 06:32 PM

Thank you very much.
Knowing my teacher, that clumsy number is probably the right one.

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