Solve the given equation $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$.
Answer
Verified610.5k+ views
Hint: For solving this question first we will find one of the roots of the equation by hit and trial and then proceed accordingly to factorise the given expression so that we can find all the roots of the given equation.
Complete step-by-step solution -
Given:
We have to find the roots of the following equation:
$12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$
Now, as the degree of the equation is 4. As such, there is not a specific method to factorise such equations, but normally we can find the roots of such equations if somehow we factorise them. But one should proceed in the right direction to factorise, so that we get the results quickly.
Now, for that when we look at the equation $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$ . Then, by hit and trial, we can say that $x=2$ is one root of the equation because when we put $x=2$ in the given equation then it satisfies it.
$\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12 \\
& \Rightarrow 12\times {{2}^{4}}-56\times {{2}^{3}}+89\times {{2}^{2}}-56\times 2+12 \\
& \Rightarrow 12\times 16-56\times 8+89\times 4-112+12 \\
& \Rightarrow 192-448+356-100 \\
& \Rightarrow 548-548 \\
& \Rightarrow 0 \\
\end{align}$
Thus, we can say that $\left( x-2 \right)$ will be one of the factors of $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12$ so, we can factorise it so, that we can take $\left( x-2 \right)$ common from each term.
Now, we will factorise $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$ and write it as $12{{x}^{4}}-24{{x}^{3}}-32{{x}^{3}}+64{{x}^{2}}+25{{x}^{2}}-50x-6x+12=0$ . Then,
$\begin{align}
& 12{{x}^{4}}-24{{x}^{3}}-32{{x}^{3}}+64{{x}^{2}}+25{{x}^{2}}-50x-6x+12=0 \\
& \Rightarrow 12{{x}^{3}}\left( x-2 \right)-32{{x}^{2}}\left( x-2 \right)+25x\left( x-2 \right)-6\left( x-2 \right)=0 \\
\end{align}$
Now, we can take $\left( x-2 \right)$ common form each term. Then,
$\begin{align}
& 12{{x}^{3}}\left( x-2 \right)-32{{x}^{2}}\left( x-2 \right)+25x\left( x-2 \right)-6\left( x-2 \right)=0 \\
& \Rightarrow \left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right)=0 \\
\end{align}$
Now, from the above result, we get, $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right)$ .
Now, we have to further factorise the terms $12{{x}^{3}}-32{{x}^{2}}+25x-6$ . Then, by hit and trial, we can say that $x=0.5$ is one of the root of the equation $12{{x}^{3}}-32{{x}^{2}}+25x-6=0$ because when we $x=0.5$ in the equation $12{{x}^{3}}-32{{x}^{2}}+25x-6=0$ then it satisfies it.
$\begin{align}
& 12{{x}^{3}}-32{{x}^{2}}+25x-6 \\
& \Rightarrow 12\times {{\left( 0.5 \right)}^{3}}-32\times {{\left( 0.5 \right)}^{2}}+25\times 0.5-6 \\
& \Rightarrow \dfrac{12}{8}-\dfrac{32}{4}+\dfrac{25}{2}-6 \\
& \Rightarrow \dfrac{3}{2}+\dfrac{25}{2}-8-6 \\
& \Rightarrow \dfrac{28}{2}-14 \\
& \Rightarrow 14-14 \\
& \Rightarrow 0 \\
\end{align}$
Thus, we can say that $\left( 2x-1 \right)$ will be one of the factors of $12{{x}^{3}}-32{{x}^{2}}+25x-6$ so, we can factorise it so, that we can take $\left( 2x-1 \right)$ common from each term.
Now, we will factorise $12{{x}^{3}}-32{{x}^{2}}+25x-6=0$ and write it as $12{{x}^{3}}-6{{x}^{2}}-26{{x}^{2}}+13x+12x-6=0$ . Then,
$\begin{align}
& 12{{x}^{3}}-6{{x}^{2}}-26{{x}^{2}}+13x+12x-6=0 \\
& \Rightarrow 6{{x}^{2}}\left( 2x-1 \right)-13x\left( 2x-1 \right)+6\left( 2x-1 \right)=0 \\
\end{align}$
Now, we can take $\left( 2x-1 \right)$ common form each term. Then,
$\begin{align}
& 6{{x}^{2}}\left( 2x-1 \right)-13x\left( 2x-1 \right)+6\left( 2x-1 \right)=0 \\
& \Rightarrow \left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right)=0 \\
\end{align}$
Now, from the above result, we get, \[12{{x}^{3}}-32{{x}^{2}}+25x-6=\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right)\] . And as we have determined that $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right)$ . Then,
\[\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right) \\
& \Rightarrow 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right) \\
\end{align}\]
Now, we have to factorise $\left( 6{{x}^{2}}-13x+6 \right)$ to find the roots and it can be done by splitting the middle term method. Then,
$\begin{align}
& 6{{x}^{2}}-13x+6 \\
& \Rightarrow 6{{x}^{2}}-9x-4x+6 \\
& \Rightarrow 3x\left( 2x-3 \right)-2\left( 2x-3 \right) \\
& \Rightarrow \left( 3x-2 \right)\left( 2x-3 \right) \\
\end{align}$
Now, from the above result, we get, $\left( 6{{x}^{2}}-13x+6 \right)=\left( 3x-2 \right)\left( 2x-3 \right)$ . And as we have determined that $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right)$ . Then,
$\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right) \\
& \Rightarrow 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 3x-2 \right)\left( 2x-3 \right) \\
\end{align}$
Now, we can easily find the roots of the $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$ . Then,
$\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0 \\
& \Rightarrow \left( x-2 \right)\left( 2x-1 \right)\left( 3x-2 \right)\left( 2x-3 \right) \\
& \Rightarrow x-2=0\text{ };\text{ }2x-1=0\text{ ; }3x-2=0\text{ ; }2x-3=0 \\
& \Rightarrow x=2\text{ ; }x=\dfrac{1}{2}\text{ ; }x=\dfrac{2}{3}\text{ ; }x=\dfrac{3}{2} \\
\end{align}$
Thus, $x=\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{2},2$ will be the roots of the given equation.
Note: Here, the student should not treat it as a tough problem it might seem tough because normally we solve quadratic equations but it is the very easy case some times which is asked so, the student should proceed as per the given methodology and just try to factorise the given term without any calculation mistake and normally for the hit and trial firstly we should try the integers like -1, 1 and 2 etc. Then, the factorisation part will become very easy and we can solve for the correct answer.
Complete step-by-step solution -
Given:
We have to find the roots of the following equation:
$12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$
Now, as the degree of the equation is 4. As such, there is not a specific method to factorise such equations, but normally we can find the roots of such equations if somehow we factorise them. But one should proceed in the right direction to factorise, so that we get the results quickly.
Now, for that when we look at the equation $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$ . Then, by hit and trial, we can say that $x=2$ is one root of the equation because when we put $x=2$ in the given equation then it satisfies it.
$\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12 \\
& \Rightarrow 12\times {{2}^{4}}-56\times {{2}^{3}}+89\times {{2}^{2}}-56\times 2+12 \\
& \Rightarrow 12\times 16-56\times 8+89\times 4-112+12 \\
& \Rightarrow 192-448+356-100 \\
& \Rightarrow 548-548 \\
& \Rightarrow 0 \\
\end{align}$
Thus, we can say that $\left( x-2 \right)$ will be one of the factors of $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12$ so, we can factorise it so, that we can take $\left( x-2 \right)$ common from each term.
Now, we will factorise $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$ and write it as $12{{x}^{4}}-24{{x}^{3}}-32{{x}^{3}}+64{{x}^{2}}+25{{x}^{2}}-50x-6x+12=0$ . Then,
$\begin{align}
& 12{{x}^{4}}-24{{x}^{3}}-32{{x}^{3}}+64{{x}^{2}}+25{{x}^{2}}-50x-6x+12=0 \\
& \Rightarrow 12{{x}^{3}}\left( x-2 \right)-32{{x}^{2}}\left( x-2 \right)+25x\left( x-2 \right)-6\left( x-2 \right)=0 \\
\end{align}$
Now, we can take $\left( x-2 \right)$ common form each term. Then,
$\begin{align}
& 12{{x}^{3}}\left( x-2 \right)-32{{x}^{2}}\left( x-2 \right)+25x\left( x-2 \right)-6\left( x-2 \right)=0 \\
& \Rightarrow \left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right)=0 \\
\end{align}$
Now, from the above result, we get, $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right)$ .
Now, we have to further factorise the terms $12{{x}^{3}}-32{{x}^{2}}+25x-6$ . Then, by hit and trial, we can say that $x=0.5$ is one of the root of the equation $12{{x}^{3}}-32{{x}^{2}}+25x-6=0$ because when we $x=0.5$ in the equation $12{{x}^{3}}-32{{x}^{2}}+25x-6=0$ then it satisfies it.
$\begin{align}
& 12{{x}^{3}}-32{{x}^{2}}+25x-6 \\
& \Rightarrow 12\times {{\left( 0.5 \right)}^{3}}-32\times {{\left( 0.5 \right)}^{2}}+25\times 0.5-6 \\
& \Rightarrow \dfrac{12}{8}-\dfrac{32}{4}+\dfrac{25}{2}-6 \\
& \Rightarrow \dfrac{3}{2}+\dfrac{25}{2}-8-6 \\
& \Rightarrow \dfrac{28}{2}-14 \\
& \Rightarrow 14-14 \\
& \Rightarrow 0 \\
\end{align}$
Thus, we can say that $\left( 2x-1 \right)$ will be one of the factors of $12{{x}^{3}}-32{{x}^{2}}+25x-6$ so, we can factorise it so, that we can take $\left( 2x-1 \right)$ common from each term.
Now, we will factorise $12{{x}^{3}}-32{{x}^{2}}+25x-6=0$ and write it as $12{{x}^{3}}-6{{x}^{2}}-26{{x}^{2}}+13x+12x-6=0$ . Then,
$\begin{align}
& 12{{x}^{3}}-6{{x}^{2}}-26{{x}^{2}}+13x+12x-6=0 \\
& \Rightarrow 6{{x}^{2}}\left( 2x-1 \right)-13x\left( 2x-1 \right)+6\left( 2x-1 \right)=0 \\
\end{align}$
Now, we can take $\left( 2x-1 \right)$ common form each term. Then,
$\begin{align}
& 6{{x}^{2}}\left( 2x-1 \right)-13x\left( 2x-1 \right)+6\left( 2x-1 \right)=0 \\
& \Rightarrow \left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right)=0 \\
\end{align}$
Now, from the above result, we get, \[12{{x}^{3}}-32{{x}^{2}}+25x-6=\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right)\] . And as we have determined that $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right)$ . Then,
\[\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 12{{x}^{3}}-32{{x}^{2}}+25x-6 \right) \\
& \Rightarrow 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right) \\
\end{align}\]
Now, we have to factorise $\left( 6{{x}^{2}}-13x+6 \right)$ to find the roots and it can be done by splitting the middle term method. Then,
$\begin{align}
& 6{{x}^{2}}-13x+6 \\
& \Rightarrow 6{{x}^{2}}-9x-4x+6 \\
& \Rightarrow 3x\left( 2x-3 \right)-2\left( 2x-3 \right) \\
& \Rightarrow \left( 3x-2 \right)\left( 2x-3 \right) \\
\end{align}$
Now, from the above result, we get, $\left( 6{{x}^{2}}-13x+6 \right)=\left( 3x-2 \right)\left( 2x-3 \right)$ . And as we have determined that $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right)$ . Then,
$\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 6{{x}^{2}}-13x+6 \right) \\
& \Rightarrow 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=\left( x-2 \right)\left( 2x-1 \right)\left( 3x-2 \right)\left( 2x-3 \right) \\
\end{align}$
Now, we can easily find the roots of the $12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0$ . Then,
$\begin{align}
& 12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0 \\
& \Rightarrow \left( x-2 \right)\left( 2x-1 \right)\left( 3x-2 \right)\left( 2x-3 \right) \\
& \Rightarrow x-2=0\text{ };\text{ }2x-1=0\text{ ; }3x-2=0\text{ ; }2x-3=0 \\
& \Rightarrow x=2\text{ ; }x=\dfrac{1}{2}\text{ ; }x=\dfrac{2}{3}\text{ ; }x=\dfrac{3}{2} \\
\end{align}$
Thus, $x=\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{2},2$ will be the roots of the given equation.
Note: Here, the student should not treat it as a tough problem it might seem tough because normally we solve quadratic equations but it is the very easy case some times which is asked so, the student should proceed as per the given methodology and just try to factorise the given term without any calculation mistake and normally for the hit and trial firstly we should try the integers like -1, 1 and 2 etc. Then, the factorisation part will become very easy and we can solve for the correct answer.
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