Answer
Verified
500.1k+ views
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.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE