The roots of [tex]3x^2-x-2=0[/tex] are [tex]\bold{1 \ or \ \frac{-2}{3}}[/tex]
Solution:
Given, [tex]3x^2-x-2=0[/tex]
The given equation is in the format [tex]ax^2+bx+c=0[/tex]
Here, a=3; b=-1; c=-2
The roots of a quadratic function are given by the quadratic formula,
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
On substituting the values we get,
[tex]x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4\times3\times(-2)}}{2\times3}[/tex]
[tex]x=\frac{+1) \pm \sqrt{1-4\times3\times(-2)}}{6}[/tex]
On solving we get,
[tex]x=\frac{+1 \pm \sqrt{1+24}}{6}[/tex]
[tex]x=\frac{+1 \pm \sqrt{25}}{6}[/tex]
Square root of 25 is 5,
[tex]x=\frac{+1+5}{6} \ or \ \frac{-1+5}{6}[/tex]
[tex]x=\frac{6}{6} \ or \ \frac{-4}{6}[/tex]
[tex]\therefore \ x=1 \ or \ \frac{-2}{3}[/tex]
