Solve the equation ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0$, two roots being 1 and 7.
Answer
Verified627.3k+ views
Hint: We will use the method of hit and trial method which is used to find the roots of the equation in an easy way. We will also use substitution here in order to select the values which satisfies the equation for which the roots will be selected.
Complete step-by-step answer:
We will consider the expression ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0$...(i). Now we will use hit and trial method to solve the expression. For this we will look at the last term which is 105 and we will use those numbers which are divisors of 105. We are already given the two roots of the equation (i) which are 1 and 7. Starting from a very small number will be a good idea here. Therefore we will first substitute the value of x = 1 which is also a divisor of 105 and we will substitute it in the left hand side of the equation (i). Therefore we will get
$\begin{align}
& {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105={{\left( 1 \right)}^{4}}-16{{\left( 1 \right)}^{3}}+86{{\left( 1 \right)}^{2}}-176\left( 1 \right)+105 \\
& \Rightarrow {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0 \\
\end{align}$
Thus we can clearly see that the value of x = 1 satisfies the expression (i). Therefore we have got one factor here which is given by (x – 1). Now we will divide the expression ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105$ by the factor (x – 1). This can be done as,
$x-1\overset{{{x}^{3}}-15{{x}^{2}}+71x-105}{\overline{\left){\begin{align}
& {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105 \\
& \underline{\pm {{x}^{4}}\mp 1{{x}^{3}}} \\
& \,\,\,\,\,\,\,-15{{x}^{3}}+86{{x}^{2}} \\
& \,\,\,\,\,\,\,\underline{\mp 15{{x}^{3}}\pm 15{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+71{{x}^{2}}-176x \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 71{{x}^{2}}\mp 71x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-105x+105 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-105x+105} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, \\
\end{align}}\right.}}$
Therefore, we have now ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105$ as $\left( x-1 \right)\left( {{x}^{3}}-15{{x}^{2}}+71x-105 \right)$. Now we will use the hit and trial method and substitute one of the smallest factors in the equation ${{x}^{3}}-15{{x}^{2}}+71x-105$ to get its factors. Now we will try solving it by substituting x = 7 in it. Therefore, we get
$\begin{align}
& {{x}^{3}}-15{{x}^{2}}+71x-105={{\left( 7 \right)}^{3}}-15{{\left( 7 \right)}^{2}}+71\left( 7 \right)-105 \\
& \Rightarrow {{x}^{3}}-15{{x}^{2}}+71x-105=343-735+497-105 \\
& \Rightarrow {{x}^{3}}-15{{x}^{2}}+71x-105=0 \\
\end{align}$
So, we can clearly see here that the value of x = 7 satisfies the equation ${{x}^{3}}-15{{x}^{2}}+71x-105=0$. Therefore we have another factor for equation (i) which is given by (x – 7). Now, we will again divide the expression ${{x}^{3}}-15{{x}^{2}}+71x-105$ by (x – 7). Therefore we get
$x-7\overset{{{x}^{2}}-8x+15}{\overline{\left){\begin{align}
& {{x}^{3}}-15{{x}^{2}}+71x-105 \\
& \underline{\pm {{x}^{3}}\mp 7{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,-8{{x}^{2}}+71x \\
& \,\,\,\,\,\,\,\,\underline{\mp 8{{x}^{2}}\pm 56x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+15x-105 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 15x\mp 105} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Thus we now have ${{x}^{3}}-15{{x}^{2}}+71x-105$ as $\left( x-7 \right)\left( {{x}^{2}}-8x+15 \right)$. Therefore we can write ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105$ as $\left( x-1 \right)\left( x-7 \right)\left( {{x}^{2}}-8x+15 \right)$. Now we will find another factor for expression (i) by substituting integers one by one. Starting from x = 3 will be great as it is also one of the factors of 15. Therefore, we have
$\begin{align}
& {{x}^{2}}-8x+15={{\left( 3 \right)}^{2}}-8\left( 3 \right)+15 \\
& \Rightarrow {{x}^{2}}-8x+15=9-24+15 \\
& \Rightarrow {{x}^{2}}-8x+15=0 \\
\end{align}$
Now we will divide the expression ${{x}^{2}}-8x+15$ by the factor (x – 3). Therefore we have
$x-3\overset{x-5}{\overline{\left){\begin{align}
& {{x}^{2}}-8x+15 \\
& \underline{\pm {{x}^{2}}\mp 3x} \\
& \,\,\,\,\,\,\,\,-5x+15 \\
& \,\,\,\,\,\,\,\,\underline{\mp 5x\pm 15} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Thus we can write ${{x}^{2}}-8x+15$ as (x – 3)(x – 5). Thus we have now converted equation (i) into its factors and since, equation (i) is equal to 0 therefore after keeping each of its factors equal to 0 w will have the values of x. This is done as,
$\begin{align}
& {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0 \\
& \Rightarrow \left( x-1 \right)\left( x-7 \right)\left( x-3 \right)\left( x-5 \right)=0 \\
\end{align}$
Hence, the values of x are 1, 7, 3 and 5.
Note: We are using here the hit and trial method and since the two roots of the equation are already given to us. Thus this method becomes much easier than any other method. When we have reached to the quadratic equation we can also use the formula to find the square root of the equation ${{x}^{2}}-8x+15$ by the formula given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This will give us the values of x directly and we will not need to do division. Since, in this question there is multiplication therefore while performing multiplication we will focus on it otherwise we can have wrong answers instead of the right one.
Complete step-by-step answer:
We will consider the expression ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0$...(i). Now we will use hit and trial method to solve the expression. For this we will look at the last term which is 105 and we will use those numbers which are divisors of 105. We are already given the two roots of the equation (i) which are 1 and 7. Starting from a very small number will be a good idea here. Therefore we will first substitute the value of x = 1 which is also a divisor of 105 and we will substitute it in the left hand side of the equation (i). Therefore we will get
$\begin{align}
& {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105={{\left( 1 \right)}^{4}}-16{{\left( 1 \right)}^{3}}+86{{\left( 1 \right)}^{2}}-176\left( 1 \right)+105 \\
& \Rightarrow {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0 \\
\end{align}$
Thus we can clearly see that the value of x = 1 satisfies the expression (i). Therefore we have got one factor here which is given by (x – 1). Now we will divide the expression ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105$ by the factor (x – 1). This can be done as,
$x-1\overset{{{x}^{3}}-15{{x}^{2}}+71x-105}{\overline{\left){\begin{align}
& {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105 \\
& \underline{\pm {{x}^{4}}\mp 1{{x}^{3}}} \\
& \,\,\,\,\,\,\,-15{{x}^{3}}+86{{x}^{2}} \\
& \,\,\,\,\,\,\,\underline{\mp 15{{x}^{3}}\pm 15{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+71{{x}^{2}}-176x \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 71{{x}^{2}}\mp 71x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-105x+105 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-105x+105} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, \\
\end{align}}\right.}}$
Therefore, we have now ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105$ as $\left( x-1 \right)\left( {{x}^{3}}-15{{x}^{2}}+71x-105 \right)$. Now we will use the hit and trial method and substitute one of the smallest factors in the equation ${{x}^{3}}-15{{x}^{2}}+71x-105$ to get its factors. Now we will try solving it by substituting x = 7 in it. Therefore, we get
$\begin{align}
& {{x}^{3}}-15{{x}^{2}}+71x-105={{\left( 7 \right)}^{3}}-15{{\left( 7 \right)}^{2}}+71\left( 7 \right)-105 \\
& \Rightarrow {{x}^{3}}-15{{x}^{2}}+71x-105=343-735+497-105 \\
& \Rightarrow {{x}^{3}}-15{{x}^{2}}+71x-105=0 \\
\end{align}$
So, we can clearly see here that the value of x = 7 satisfies the equation ${{x}^{3}}-15{{x}^{2}}+71x-105=0$. Therefore we have another factor for equation (i) which is given by (x – 7). Now, we will again divide the expression ${{x}^{3}}-15{{x}^{2}}+71x-105$ by (x – 7). Therefore we get
$x-7\overset{{{x}^{2}}-8x+15}{\overline{\left){\begin{align}
& {{x}^{3}}-15{{x}^{2}}+71x-105 \\
& \underline{\pm {{x}^{3}}\mp 7{{x}^{2}}} \\
& \,\,\,\,\,\,\,\,-8{{x}^{2}}+71x \\
& \,\,\,\,\,\,\,\,\underline{\mp 8{{x}^{2}}\pm 56x} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+15x-105 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\pm 15x\mp 105} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Thus we now have ${{x}^{3}}-15{{x}^{2}}+71x-105$ as $\left( x-7 \right)\left( {{x}^{2}}-8x+15 \right)$. Therefore we can write ${{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105$ as $\left( x-1 \right)\left( x-7 \right)\left( {{x}^{2}}-8x+15 \right)$. Now we will find another factor for expression (i) by substituting integers one by one. Starting from x = 3 will be great as it is also one of the factors of 15. Therefore, we have
$\begin{align}
& {{x}^{2}}-8x+15={{\left( 3 \right)}^{2}}-8\left( 3 \right)+15 \\
& \Rightarrow {{x}^{2}}-8x+15=9-24+15 \\
& \Rightarrow {{x}^{2}}-8x+15=0 \\
\end{align}$
Now we will divide the expression ${{x}^{2}}-8x+15$ by the factor (x – 3). Therefore we have
$x-3\overset{x-5}{\overline{\left){\begin{align}
& {{x}^{2}}-8x+15 \\
& \underline{\pm {{x}^{2}}\mp 3x} \\
& \,\,\,\,\,\,\,\,-5x+15 \\
& \,\,\,\,\,\,\,\,\underline{\mp 5x\pm 15} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}}\right.}}$
Thus we can write ${{x}^{2}}-8x+15$ as (x – 3)(x – 5). Thus we have now converted equation (i) into its factors and since, equation (i) is equal to 0 therefore after keeping each of its factors equal to 0 w will have the values of x. This is done as,
$\begin{align}
& {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0 \\
& \Rightarrow \left( x-1 \right)\left( x-7 \right)\left( x-3 \right)\left( x-5 \right)=0 \\
\end{align}$
Hence, the values of x are 1, 7, 3 and 5.
Note: We are using here the hit and trial method and since the two roots of the equation are already given to us. Thus this method becomes much easier than any other method. When we have reached to the quadratic equation we can also use the formula to find the square root of the equation ${{x}^{2}}-8x+15$ by the formula given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This will give us the values of x directly and we will not need to do division. Since, in this question there is multiplication therefore while performing multiplication we will focus on it otherwise we can have wrong answers instead of the right one.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 10 General Knowledge: Engaging Questions & Answers for Success
Master Class 10 Computer Science: Engaging Questions & Answers for Success
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 Science: Engaging Questions & Answers for Success
Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
In cricket, what is the term for a bowler taking five wickets in an innings?
Who Won 36 Oscar Awards? Record Holder Revealed
What is the name of Japan Parliament?
What is the median of the first 10 natural numbers class 10 maths CBSE
Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE