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# Solve the following equation: $3{{x}^{2}}-7+3\sqrt{3{{x}^{2}}-16x+21}=16x$ Verified
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Hint: To solve the given equation, replace the term $3{{x}^{2}}-16x+21$ by ${{a}^{2}}$ and then rearrange the terms to factorize them into product of linear terms and find solution of the given equation.

We have the equation $3{{x}^{2}}-7+3\sqrt{3{{x}^{2}}-16x+21}=16x$. We have to solve this equation.
To begin with, we will replace the term $3{{x}^{2}}-16x+21$by${{a}^{2}}$.
$\Rightarrow 3{{x}^{2}}-16x+21={{a}^{2}}$ $...\left( 1 \right)$
We can rearrange the terms and write the above equation as $3{{x}^{2}}-16x={{a}^{2}}-21$. $...\left( 2 \right)$
Substituting equation $\left( 1 \right)$and$\left( 2 \right)$ in the equation $3{{x}^{2}}-7+3\sqrt{3{{x}^{2}}-16x+21}-16x=0$, we have ${{a}^{2}}-21-7+3\sqrt{{{a}^{2}}}=0$.
$\Rightarrow {{a}^{2}}-28+3a=0$
Now, we will factorize the above equation.
Thus, we have ${{a}^{2}}+7a-4a-28=0$.
\begin{align} & \Rightarrow a\left( a+7 \right)-4\left( a+7 \right)=0 \\ & \Rightarrow \left( a-4 \right)\left( a+7 \right)=0 \\ & \Rightarrow a=4,a=-7 \\ \end{align}
We will now substitute the values $a=4,a=-7$ in equation$\left( 1 \right)$,
Substituting $a=4$, we have $3{{x}^{2}}-16x+21={{4}^{2}}$.
\begin{align} & \Rightarrow 3{{x}^{2}}-16x+21=16 \\ & \Rightarrow 3{{x}^{2}}-16x+5=0 \\ & \Rightarrow 3{{x}^{2}}-x-15x+5=0 \\ & \Rightarrow x\left( 3x-1 \right)-5\left( 3x-1 \right)=0 \\ & \Rightarrow \left( x-5 \right)\left( 3x-1 \right)=0 \\ & \Rightarrow x=5,\dfrac{1}{3} \\ \end{align}
Substituting $a=-7$, we have $3{{x}^{2}}-16x+21={{\left( -7 \right)}^{2}}$.
\begin{align} & \Rightarrow 3{{x}^{2}}-16x+21=49 \\ & \Rightarrow 3{{x}^{2}}-16x+-28=0 \\ \end{align}
As we can’t easily factorize the above equation to find the roots, we will use the formula which states that the roots of the equation $a{{x}^{2}}+bx+c=0$ are of the form $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Substituting $a=3,b=-16,c=-28$ in the above equation, we have roots of the equation$3{{x}^{2}}-16x-28=0$ of the form $x=\dfrac{16\pm \sqrt{{{\left( -16 \right)}^{2}}-4\left( 3 \right)\left( -28 \right)}}{2\left( 3 \right)}=\dfrac{16\pm \sqrt{592}}{6}=\dfrac{16\pm 4\sqrt{37}}{6}=\dfrac{8\pm 2\sqrt{37}}{3}$.
Hence, by solving the equation $3{{x}^{2}}-7+3\sqrt{3{{x}^{2}}-16x+21}=16x$, we get $x=5,\dfrac{1}{3},\dfrac{8+2\sqrt{37}}{3},\dfrac{8-2\sqrt{37}}{3}$ as the roots of the equation.
We observe that the equation given to us is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables.
Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree $4$. This is because when we rearrange the terms and square the equation on both sides, we get a polynomial whose highest degree of terms is $4$.

Note: We can also solve this question by rearranging the terms and then squaring the equation on both sides to remove the square root from the equation.
Last updated date: 25th Sep 2023
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