Solve the equation ${{x}^{4}}-11{{x}^{3}}+44{{x}^{2}}-76x+48=0$ which has equal roots.
Answer
Verified599.4k+ views
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE
Find the coordinates of the midpoint of the line segment class 11 maths CBSE
Distinguish between static friction limiting friction class 11 physics CBSE
The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE
The first National Commission on Labour NCL submitted class 11 social science CBSE
Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE
Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE
State the laws of reflection of light