Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
$3 x^{2}+11 x+10=0$
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 $-\frac{5}{3}$.
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