Find the roots of the following quadratic equations

Solution:

We have been given that,

$3 x^{2}+11 x+10=0$

Now divide throughout by 3. We get,

$x^{2}+\frac{11}{3} x+\frac{10}{3}=0$

Now take the constant term to the RHS and we get

$x^{2}+\frac{11}{3} x=-\frac{10}{3}$

Now add square of half of co-efficient of ‘x’ on both the sides. We have,

$x^{2}+\frac{11}{3} x+\left(\frac{11}{6}\right)^{2}=\left(\frac{11}{6}\right)^{2}-\frac{10}{3}$

$x^{2}+\left(\frac{11}{6}\right)^{2}+2\left(\frac{11}{3}\right) x=\frac{1}{36}$

$\left(x+\frac{11}{6}\right)^{2}=\frac{1}{36}$

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

$x+\frac{11}{6}=\pm \frac{1}{6}$

$x=-\frac{11}{6} \pm \frac{1}{6}$

Now, we have the values of ‘x’ as

$x=-\frac{11}{6}+\frac{1}{6}$

$=-\frac{5}{3}$

Also we have,

$x=-\frac{11}{6}-\frac{1}{6}$

$=-2$

Therefore the roots of the equation are $-2$ and $-\frac{5}{3}$.