Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

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

Given:

Given quadratic equation is $3x^2+11x + 10 = 0$.

To do:

We have to find the roots of the given quadratic equation.

Solution:

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

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

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

$x^2 + 2\times \frac{1}{2}\times \frac{11}{3} x = -\frac{10}{3}$  

$x^2 + 2\times \frac{11}{6} x = -\frac{10}{3}$  

Adding $(\frac{11}{6})^2$ on both sides completes the square. Therefore,

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

$(x+\frac{11}{6})^2=-\frac{10}{3}+\frac{121}{36}$      (Since $(a+b)^2=a^2+2ab+b^2$)

$(x+\frac{11}{6})^2=\frac{121-12\times10}{36}$

$(x+\frac{11}{6})^2=\frac{121-120}{36}$

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

$x+\frac{11}{6}=\pm \sqrt{\frac{1}{36}}$     (Taking square root on both sides)

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

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

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

$x=\frac{-10}{6}$ or $x=-\frac{12}{6}$

$x=-\frac{5}{3}$ or $x=-2$

The values of $x$ are $-2$ and $-\frac{5}{3}$.

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Updated on: 10-Oct-2022

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