Question

# Solve the following equation:$\sqrt{3{{x}^{2}}-2x+9}+\sqrt{3{{x}^{2}}-2x-4}=13$

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Hint: In this question, we first need to take one of the square root term to the other side and then do squaring on both sides. Then cancel the common terms on both sides and rearrange them and again do squaring on both sides. On further simplification we shall get the result.

From the given equation in the question we have,
$\Rightarrow \sqrt{3{{x}^{2}}-2x+9}+\sqrt{3{{x}^{2}}-2x-4}=13$
Now, by taking one of the square root term to the other side we get,
$\Rightarrow \sqrt{3{{x}^{2}}-2x+9}=13-\sqrt{3{{x}^{2}}-2x-4}$
Let us now do the squaring on both sides to the above equation.
$\Rightarrow {{\left( \sqrt{3{{x}^{2}}-2x+9} \right)}^{2}}={{\left( 13-\sqrt{3{{x}^{2}}-2x-4} \right)}^{2}}$
Now, on squaring and expanding the terms on both sides Using $(a-b)^2$=$a^2+b^2-2ab$ formula further we get,
$\Rightarrow 3{{x}^{2}}-2x+9={{\left( 13 \right)}^{2}}+3{{x}^{2}}-2x-4-2\times 13\times \sqrt{3{{x}^{2}}-2x-4}$
Now, on cancelling the common terms on both sides and rearranging we get,
$\Rightarrow 9+4-169=-26\times \sqrt{3{{x}^{2}}-2x-4}$
Now, on further simplification we get,
$\Rightarrow -156=-26\times \sqrt{3{{x}^{2}}-2x-4}$
Let us now divide with -26 on both the sides then we get,
$\Rightarrow 6=\sqrt{3{{x}^{2}}-2x-4}$
Now, again by squaring on both sides we get,
$\Rightarrow {{\left( 6 \right)}^{2}}={{\left( \sqrt{3{{x}^{2}}-2x-4} \right)}^{2}}$
From this, we can rewrite as.
$\Rightarrow 3{{x}^{2}}-2x-4=36$
Now, on rearranging the terms we get a quadratic equation in x.
$\Rightarrow 3{{x}^{2}}-2x-40=0$
Now, this can be further solved by using the direct formula to get the value of x.
As we already know that quadratic equation $a{{x}^{2}}+bx+c=0$ has two roots, given by
$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, on comparing the quadratic equation we got with the above one we get,
$a=3,b=-2,c=-40$
Now, by substituting these values in the above direct formula we get,
$\Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 3\times \left( -40 \right)}}{2\times 3}$
Now, by simplifying this further we get,
\begin{align} & \Rightarrow x=\dfrac{2\pm \sqrt{4+480}}{6} \\ & \Rightarrow x=\dfrac{2\pm \sqrt{484}}{6} \\ & \Rightarrow x=\dfrac{2\pm 22}{6} \\ \end{align}
Here, we have two possible values of x assuming as x1 and x2
\begin{align} & \Rightarrow {{x}_{1}}=\dfrac{2+22}{6} \\ & \Rightarrow {{x}_{1}}=\dfrac{24}{6} \\ & \therefore {{x}_{1}}=4 \\ \end{align}
\begin{align} & \Rightarrow {{x}_{2}}=\dfrac{2-22}{6} \\ & \Rightarrow {{x}_{2}}=\dfrac{-20}{6} \\ & \therefore {{x}_{2}}=\dfrac{-10}{3} \\ \end{align}
Hence the values of x that satisfy the given equation are 4, $\dfrac{-10}{3}$

Note: Instead of taking one of the square root terms to the other side and then squaring on both sides we can also solve the equation by directly squaring on both sides and then again rearrange the terms and then do squaring on both sides. This results in the formation of bi-quadratic equation which is difficult to solve.
It is important to note that the value of x can be found by using the factorisation method instead of the direct formula but it takes a lot of time to check the possible values that satisfy the given equation by trial and error method.