# Thread: Solving Quadratic Equations Using The Square Root Principal

1. ## Solving Quadratic Equations Using The Square Root Principal

I need help with this question:$\displaystyle 5/2 = 6/(x+4)^2$ (Rationalize the denominator) Thanks

2. The denominators already are rational.

You can cross multiply then solve the equation. If that's what you want to do?

$\displaystyle \frac{5}{2}=\frac{6}{(x+4)^2}$

$\displaystyle 5(x+4)^2=12$

Can you take it from there?

3. Originally Posted by Stroodle
The denominators already are rational.

You can cross multiply then solve the equation. If that's what you want to do?

$\displaystyle \frac{5}{2}=\frac{6}{(x+4)^2}$

$\displaystyle 5(x+4)^2=12$

Can you take it from there?
Oh Wait I was FOILing $\displaystyle 5(x+4)^2$ instead of taking the square root, thanks. So after that i get $\displaystyle 5x+20 = 2sqrt3$ subtract 20 and divide by 5 to get$\displaystyle x = (2sqrt3)/5 - 4$ Is that right?

4. $\displaystyle 5(x+4)^2=12$

$\displaystyle 5(x+4)(x+4)=12$

$\displaystyle 5(x^2+8x+16)=12$

$\displaystyle 5x^2+40x+68=0$

Now you can either sub these values into the quadratic formula or complete the square to find the values of x.

Do you know how to do this?

5. I would use the quadratic formula but it says solve using the square root principal, is that just isolating x?

6. Oh, sorry I misunderstood. I'm not familiar with the square root principle. But I assume it's:

$\displaystyle 5(x+4)^2=12$

$\displaystyle (x+4)^2=\frac{12}{5}$

$\displaystyle x+4=\pm\sqrt{\frac{12}{5}}$

$\displaystyle x=\pm\sqrt{\frac{12}{5}}-4$

7. Yes that is it but you have to rationalize the denominator so would the final answer be $\displaystyle x=2sqrt(15)/5-4?$

8. Originally Posted by Stroodle
Oh, sorry I misunderstood. I'm not familiar with the square root principle. But I assume it's:

$\displaystyle 5(x+4)^2=12$

$\displaystyle (x+4)^2=\frac{12}{5}$

$\displaystyle x+4=\pm\sqrt{\frac{12}{5}}$

$\displaystyle x=\pm\sqrt{\frac{12}{5}}-4$
I'd make sure to rationalise the denominator...

$\displaystyle 5(x + 4)^2 = 12$

$\displaystyle (x + 4)^2 = \frac{12}{5}$

$\displaystyle x + 4 = \pm\sqrt{\frac{12}{5}}$

$\displaystyle x + 4 = \pm \frac{\sqrt{12}}{\sqrt{5}}$

$\displaystyle x + 4 = \pm \frac{2\sqrt{3}}{\sqrt{5}}$

$\displaystyle x + 4 = \pm \frac{2\sqrt{15}}{5}$

$\displaystyle x = -4 \pm \frac{2\sqrt{15}}{5}$

$\displaystyle x = \frac{-20 \pm \sqrt{15}}{5}$.

This should be the same answer you get if you use the Quadratic Formula...

9. Yep. The answer would be $\displaystyle x=\pm\frac{2\sqrt{15}}{5}-4$

10. Oh yea plus or minus, thanks for your help I really appreciate it.

11. No probs. Finally got there in the end