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

$\sqrt3x^2 +10x + 7\sqrt3 = 0$

Given:

Given quadratic equation is $\sqrt3x^2 +10x + 7\sqrt3 = 0$.

To do:

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

Solution:

$\sqrt3x^2 +10x + 7\sqrt3 = 0$

$\sqrt3(x^2 +\frac{10}{\sqrt3} x +7)=0$  

$x^2+\frac{10}{\sqrt3}x+7=0$

$x^2+2\times \frac{1}{2} \times \frac{10}{\sqrt3} x =-7$

$x^2+2(\frac{10}{2\sqrt3})x=-7$

$x^2+2(\frac{5}{\sqrt3})x=-7$

Adding $(\frac{5}{\sqrt3})^2$ on both sides completes the square. Therefore,

$x^2+2(\frac{5}{\sqrt3})x+(\frac{5}{\sqrt3})^2=-7+(\frac{5}{\sqrt3})^2$

$(x+\frac{5}{\sqrt3})^2=-7+\frac{25}{3}$      (Since $(a+b)^2=a^2+2ab+b^2$)

$(x+\frac{5}{\sqrt3})^2=\frac{25-7\times3}{3}$

$(x+\frac{5}{\sqrt3})^2=\frac{25-21}{8}$

$x+\frac{5}{\sqrt3}=\pm \sqrt{\frac{4}{3}}$

$x+\frac{5}{\sqrt3}=\pm \frac{2}{\sqrt3}$

$x=-\frac{5}{\sqrt3}+\frac{2}{\sqrt3}$ or $x=-\frac{5}{\sqrt3}-\frac{2}{\sqrt3}$

$x=\frac{2-5}{\sqrt3}$ or $x=-(\frac{5+2}{\sqrt3})$

$x=\frac{-3}{\sqrt3}$ or $x=-(\frac{7}{\sqrt3})$

$x=-\sqrt3$ or $x=-\frac{7}{\sqrt3}$

The values of $x$ are $-\sqrt3$ and $-\frac{7}{\sqrt3}$.

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Updated on: 10-Oct-2022

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