The answer to the question below is x = 7. How does one get 7 in this sample question?

sqrt{5x + 1} + 3x = 27

2. Originally Posted by magentarita
The answer to the question below is x = 7. How does one get 7 in this sample question?

sqrt{5x + 1} + 3x = 27
$\displaystyle \sqrt{5x+1}+3x=27$
$\displaystyle \sqrt{5x+1}=27-3x$
Square both sides.
$\displaystyle 5x+1=9x^2-162x+729$
$\displaystyle 9x^2-167x+728=0$
$\displaystyle a=9$ $\displaystyle b=-167$ $\displaystyle c=728$
Use the quadratic formula:$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
And you end up with $\displaystyle x=7$ or $\displaystyle x=\frac{104}{9}$

One slight problem there is that 104/9 is completely wrong as you can tell by putting it in. However, 7 fits perfectly and is the only real solution as far as I can tell.

3. ## yes...

Originally Posted by JeWiSh
$\displaystyle \sqrt{5x+1}+3x=27$
$\displaystyle \sqrt{5x+1}=27-3x$
Square both sides.
$\displaystyle 5x+1=9x^2-162x+729$
$\displaystyle 9x^2-167x+728=0$
$\displaystyle a=9$ $\displaystyle b=-167$ $\displaystyle c=728$
Use the quadratic formula:$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
And you end up with $\displaystyle x=7$ or $\displaystyle x=\frac{104}{9}$

One slight problem there is that 104/9 is completely wrong as you can tell by putting it in. However, 7 fits perfectly and is the only real solution as far as I can tell.
I was going to use the quadratic formula but decided not to.