ex1.5_13

$x$ $\quad 3x^2-8x+1=0$

If $(1)(3)=3$ $-8$ . There are no such integers, so we give up on factoring. Let's go right to the quadratic formula.

In this problem, we have

a=3, \, b=-8, \, c=1.

It follows that

x = \frac{-(-8) \pm \sqrt{(-8)^2-4(3)(1)}}{2(3)}

x = \frac{8 \pm \sqrt{52}}{6} = \frac{8 \pm 2 \sqrt{13}}{6}.

There are two solutions,

x=\frac{8+2\sqrt{13}}{6} \quad \mbox{or} \quad x=\frac{8-2\sqrt{13}}{6},

$2$ . After doing so, we have

x=\frac{4+\sqrt{13}}{3} \quad \mbox{or} \quad x=\frac{4-\sqrt{13}}{3}.