Question

Find the exact solutions and simplify as much as possible: 3x^2+1=6x

Answer 1

Answer:

$\frac{3+ \sqrt{6} }{3}$ and $\frac{3- \sqrt{6} }{3}$

Step-by-step explanation:

This is a quadratic equation, so we start by rearranging its terms so we can apply the quadratic formula:

$ax^2+bx+c=0$ has two possible solutions given by:

$x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

So we rearrange the terms in the given quadratic equation bringing all terms to the left hhand side of the equal sign:

$3x^2+1=6x\\3x^2-6x+1=0$

The solutions will then be:

$x=\frac{6+/- \sqrt{36-4(3)(1)} }{2(3)}=\frac{6+/- \sqrt{36-12} }{6}=\frac{6+/- \sqrt{24} }{6}=\frac{6+/- \sqrt{4*6} }{6}=\frac{6+/- 2\sqrt{6} }{6}=\frac{3+/- \sqrt{6} }{3}$

Therefore the two possible answer are: $\frac{3+ \sqrt{6} }{3}$ and $\frac{3- \sqrt{6} }{3}$