Solve the equation ${{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48=0$ which has equal roots.
Answer
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Hint: We will use the hit and trial method to solve the equation. In this method, we substitute the value of x and we collect those values of x which satisfies the equation or polynomial of any degree.
Complete step-by-step answer:
We will consider the polynomial given by ${{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48=0...(i)$. Now, we will use the hit and trial method. In this method we will substitute the value of x to any integer and select those values of x which satisfies the equation and polynomial. Here, we are given a fourth degree polynomial. We will substitute the value of x = 0 in equation (i). Therefore, we have
${{\left( 0 \right)}^{4}}-11{{\left( 0 \right)}^{3}}+44{{\left( 0 \right)}^{2}}-76\left( 0 \right)+48=0$
Since, $48\ne 0$ therefore x = 0 cannot be a factor of (i). Now, we will substitute x = 2 in equation (i). Therefore, we have
$\begin{align}
& {{\left( 2 \right)}^{4}}-11{{\left( 2 \right)}^{3}}+44{{\left( 2 \right)}^{2}}-76\left( 2 \right)+48=16-88+176-152+48 \\
& \Rightarrow 16-88+176-152+48=0 \\
\end{align}$
Since, x = 2 satisfies the equation (i). And now we will divide the equation (i) by (x - 2) where (x - 2) is a factor of equation (i). Therefore, we have
$x-2\overset{{{x}^{3}}-9{{x}^{2}}+26x-24}{\overline{\left){\begin{align}
& {{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48 \\
& \underline{\pm {{x}^{4}}\mp 2{{x}^{3}}} \\
& \,\,\,\,\,\,\,\,-9{{x}^{3}}+44{{x}^{2}} \\
& \,\,\,\,\,\,\,\,\underline{\mp 9{{x}^{3}}\pm 18{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+26{{x}^{2}}-76x \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 26{{x}^{2}}\mp 52x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-24x+48 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\mp 24x\pm 48} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Thus, our equation ${{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48=0$ is factored into $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=0$. Now, we will consider the hit and trial method again to the equation ${{x}^{3}}-9{{x}^{2}}+26x-24=0$ as $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=0$ can also be written as x - 2 = 0 and ${{x}^{3}}-9{{x}^{2}}+26x-24=0$. So, by substituting the value of x = 2 again in the equation ${{x}^{3}}-9{{x}^{2}}+26x-24=0$ we get,
$\begin{align}
& {{\left( 2 \right)}^{3}}-9{{\left( 2 \right)}^{2}}+26\left( 2 \right)-24=8-36+52-24 \\
& \Rightarrow 8-36+52-24=0 \\
\end{align}$
Therefore, we have one more factor which is (x - 2). Now, we can write $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=0$ as a new factor. But for that we will divide by x - 2. Thus, we have
$x-2\overset{{{x}^{2}}-7x+12}{\overline{\left){\begin{align}
& {{x}^{3}}-9{{x}^{2}}+26x-24 \\
& \underline{\pm {{x}^{3}}\mp 2{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,-7{{x}^{2}}+26x \\
& \,\,\,\,\,\,\,\,\underline{\mp 7{{x}^{2}}\pm 14x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+12x-24 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 12x\mp 24} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Therefore, we have $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=\left( x-2 \right)\left( x-2 \right)\left( {{x}^{2}}-7x+12 \right)$. Now, we will use the hit and trial method on the equation ${{x}^{2}}-7x+12=0$. By substituting x = 3 in this equation we have,
$\begin{align}
& {{\left( 3 \right)}^{2}}-7\left( 3 \right)+12=9-21+12 \\
& 9-21+12=0 \\
\end{align}$
So, we have another root (x - 3). So, we can solve the equation by dividing ${{x}^{2}}-7x+12=0$ by (x - 3). Therefore, we have
$x-3\overset{x-4}{\overline{\left){\begin{align}
& {{x}^{2}}-7x+12 \\
& \underline{\pm {{x}^{2}}\mp 3x} \\
& \,\,\,\,\,\,\,\,\,-4x+12 \\
& \,\,\,\,\,\,\,\,\,\underline{\mp 4x+12} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Therefore, we have $\left( x-2 \right)\left( x-2 \right)\left( {{x}^{2}}-7x+12 \right)=\left( x-2 \right)\left( x-2 \right)\left( x-3 \right)\left( x-4 \right)$.
So, by keeping this equation to 0 we have x = 2, 2, 3, 4. And clearly, this equation (i) has two equal roots.
Hence, the required values of equation (i) are 2, 2, 3, 4.
Note: We could have solved the equation ${{x}^{2}}-7x+12=0$ by using the formula of square roots alternatively. Now we can use this formula for the equation of the form $a{{x}^{2}}+bx+c=0$. And the roots are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. While solving equations of such high degrees like 4th degree polynomial results into a complex solution. Here, hit and trial method works the best.
Complete step-by-step answer:
We will consider the polynomial given by ${{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48=0...(i)$. Now, we will use the hit and trial method. In this method we will substitute the value of x to any integer and select those values of x which satisfies the equation and polynomial. Here, we are given a fourth degree polynomial. We will substitute the value of x = 0 in equation (i). Therefore, we have
${{\left( 0 \right)}^{4}}-11{{\left( 0 \right)}^{3}}+44{{\left( 0 \right)}^{2}}-76\left( 0 \right)+48=0$
Since, $48\ne 0$ therefore x = 0 cannot be a factor of (i). Now, we will substitute x = 2 in equation (i). Therefore, we have
$\begin{align}
& {{\left( 2 \right)}^{4}}-11{{\left( 2 \right)}^{3}}+44{{\left( 2 \right)}^{2}}-76\left( 2 \right)+48=16-88+176-152+48 \\
& \Rightarrow 16-88+176-152+48=0 \\
\end{align}$
Since, x = 2 satisfies the equation (i). And now we will divide the equation (i) by (x - 2) where (x - 2) is a factor of equation (i). Therefore, we have
$x-2\overset{{{x}^{3}}-9{{x}^{2}}+26x-24}{\overline{\left){\begin{align}
& {{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48 \\
& \underline{\pm {{x}^{4}}\mp 2{{x}^{3}}} \\
& \,\,\,\,\,\,\,\,-9{{x}^{3}}+44{{x}^{2}} \\
& \,\,\,\,\,\,\,\,\underline{\mp 9{{x}^{3}}\pm 18{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+26{{x}^{2}}-76x \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 26{{x}^{2}}\mp 52x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-24x+48 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\mp 24x\pm 48} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Thus, our equation ${{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48=0$ is factored into $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=0$. Now, we will consider the hit and trial method again to the equation ${{x}^{3}}-9{{x}^{2}}+26x-24=0$ as $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=0$ can also be written as x - 2 = 0 and ${{x}^{3}}-9{{x}^{2}}+26x-24=0$. So, by substituting the value of x = 2 again in the equation ${{x}^{3}}-9{{x}^{2}}+26x-24=0$ we get,
$\begin{align}
& {{\left( 2 \right)}^{3}}-9{{\left( 2 \right)}^{2}}+26\left( 2 \right)-24=8-36+52-24 \\
& \Rightarrow 8-36+52-24=0 \\
\end{align}$
Therefore, we have one more factor which is (x - 2). Now, we can write $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=0$ as a new factor. But for that we will divide by x - 2. Thus, we have
$x-2\overset{{{x}^{2}}-7x+12}{\overline{\left){\begin{align}
& {{x}^{3}}-9{{x}^{2}}+26x-24 \\
& \underline{\pm {{x}^{3}}\mp 2{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,-7{{x}^{2}}+26x \\
& \,\,\,\,\,\,\,\,\underline{\mp 7{{x}^{2}}\pm 14x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+12x-24 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 12x\mp 24} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Therefore, we have $\left( x-2 \right)\left( {{x}^{3}}-9{{x}^{2}}+26x-24 \right)=\left( x-2 \right)\left( x-2 \right)\left( {{x}^{2}}-7x+12 \right)$. Now, we will use the hit and trial method on the equation ${{x}^{2}}-7x+12=0$. By substituting x = 3 in this equation we have,
$\begin{align}
& {{\left( 3 \right)}^{2}}-7\left( 3 \right)+12=9-21+12 \\
& 9-21+12=0 \\
\end{align}$
So, we have another root (x - 3). So, we can solve the equation by dividing ${{x}^{2}}-7x+12=0$ by (x - 3). Therefore, we have
$x-3\overset{x-4}{\overline{\left){\begin{align}
& {{x}^{2}}-7x+12 \\
& \underline{\pm {{x}^{2}}\mp 3x} \\
& \,\,\,\,\,\,\,\,\,-4x+12 \\
& \,\,\,\,\,\,\,\,\,\underline{\mp 4x+12} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Therefore, we have $\left( x-2 \right)\left( x-2 \right)\left( {{x}^{2}}-7x+12 \right)=\left( x-2 \right)\left( x-2 \right)\left( x-3 \right)\left( x-4 \right)$.
So, by keeping this equation to 0 we have x = 2, 2, 3, 4. And clearly, this equation (i) has two equal roots.
Hence, the required values of equation (i) are 2, 2, 3, 4.
Note: We could have solved the equation ${{x}^{2}}-7x+12=0$ by using the formula of square roots alternatively. Now we can use this formula for the equation of the form $a{{x}^{2}}+bx+c=0$. And the roots are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. While solving equations of such high degrees like 4th degree polynomial results into a complex solution. Here, hit and trial method works the best.
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