# Thread: Solving for the square roots of binomials

1. ## Solving for the square roots of binomials

I have no idea how to do this one:

Sqrt(3x+4) + Sqrt(x-3) = 5

I did guess and check and found that x = 4, but I want a more elegant solution. I did try squaring both sides and came up with

3x + 4 + x - 3 + 2Sqrt(3x^2 - 5x -12) = 25 but, long story short, that did not turn out right for me. Am i on the right track?

2. Originally Posted by satx
I have no idea how to do this one:

Sqrt(3x+4) + Sqrt(x-3) = 5

I did guess and check and found that x = 4, but I want a more elegant solution. I did try squaring both sides and came up with

3x + 4 + x - 3 + 2Sqrt(3x^2 - 5x -12) = 25 but, long story short, that did not turn out right for me. Am i on the right track?
HI

Yes , you will need to start by squaring .

$\displaystyle [\sqrt{3x+4}+\sqrt{x-3}]^2=[5]^2$

$\displaystyle 4x+1+2\sqrt{3x^2-5x-12}=25$

$\displaystyle 12-2x=\sqrt{3x^2-5x-12}$

Then square both sides once again and simplify .

$\displaystyle x^2-43x+156=0$

$\displaystyle (x-4)(x-39)=0$

x=4 , 39

Now test these 2 values . Only 4 is valid .

3. cheers mate

4. Hello, satx!

Another approach . . .

$\displaystyle \sqrt{3x+4} + \sqrt{x-3} \:=\: 5$

. . $\displaystyle \sqrt{3x+4} \;=\;5-\sqrt{x-3}$

Square both sides: .$\displaystyle (\sqrt{3x+4})^2 \;=\;(5-\sqrt{x-3})^2 \quad\Rightarrow\quad 3x + 4 \;=\;25 - 10\sqrt{x-3} + x-3$

. . $\displaystyle 10\sqrt{x-3} \;=\;18 - 2x \quad\Rightarrow\quad 5\sqrt{x-3} \;=\;9-x$

Square both sides: .$\displaystyle (5\sqrt{x-3})^2 \;=\;(9-x)^2 \quad\Rightarrow\quad 25(x-3) \;=\;81 - 18x + x^2$

. . $\displaystyle x^2 - 43x + 156 \:=\:0 \quad\Rightarrow\quad (x-4)(x-39) \:=\:0 \quad\rightarrow\quad x \:=\:4,\:39$

As mathaddict pointed out, $\displaystyle x = 39$ is extraneous.

. . Therefore, only $\displaystyle x = 4$ is a solution.