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?
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.
Hello, yewchung!
I tried the problem and got clumsy answers, but they checked out.
(Well, one of them checked . . . the other was extraneous.)
$\displaystyle \text{Solve for }x\!:\;\sqrt{x+5} - \sqrt{2x+3} \:=\: -2$
Isolate a radical: .$\displaystyle \sqrt{2x+3} \:=\:\sqrt{x+5} + 2$
Square both sides: .$\displaystyle \left(\sqrt{2x+3}\right)^2 \;=\;\left(\sqrt{x+5}+2\right)^2$
n . . . . . . . . . . . . . . . . . $\displaystyle 2x + 3 \;=\;x+5 + 4\sqrt{x+5} + 4$
Isolate the radical: . . . . . $\displaystyle x - 6 \;=\;4\sqrt{x+5}$
Square both sides: . . . $\displaystyle (x -6)^2 \;=\;\left(4\sqrt{x+5}\right)^2$
n . . . . . . . . . . . $\displaystyle x^2 - 12x + 36 \;=\;16(x+5)$
n . . . . . . . . . . . $\displaystyle x^2 - 12x + 36 \;=\;16x + 80$
n . . . . . . . . . . . $\displaystyle x^2 - 28x - 44 \;=\;0$
Quadratic Formula: . $\displaystyle x\;=\;\frac{28\pm\sqrt{960}}{2} \;=\;\frac{28\pm8\sqrt{15}}{2} \;=\;14 \pm4\sqrt{15}$
The only root is: .$\displaystyle x \;=\;14 + 4\sqrt{15} \;=\;29.491933385\hdots$