# Solve the following equation: \[3{{x}^{2}}-7+3\sqrt{3{{x}^{2}}-16x+21}=16x\]

Last updated date: 23rd Mar 2023

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Answer

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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.

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.

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