
The transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is
(A) ${x^4} - 24{x^2} + 65x - 55 = 0$
(B) ${x^4} + 24{x^2} + 65x + 55 = 0$
(C) ${x^4} - 24{x^2} - 65x + 55 = 0$
(D) ${x^4} + 24{x^2} + 65x - 55 = 0$
Answer
555k+ views
Hint: At first use the substitution \[\left( {x + h} \right)\] in the place of x in the given equation. Now note that, in order to eliminate the second term from this transformed equation, the coefficient of \[{x^3}\] must be zero. Therefore find h and rewrite the transformed equation substituting the value of h.
Complete step by step answer:
According to the question we are to eliminate the second term from the equation ${x^4} + 8{x^3} + x - 5 = 0$..... (1)
Let us at first use the substitution \[\left( {x + h} \right)\] in the place of x in equation (1).
Therefore we have,
${\left( {x + h} \right)^4} + 8{\left( {x + h} \right)^3} + \left( {x + h} \right) - 5 = 0$
On expanding we get,
$ \Rightarrow {x^4} + 4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4} + 8\left[ {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right] + x + h - 5 = 0$
On taking terms common we get,
$ \Rightarrow {x^4} + \left( {4h + 8} \right){x^3} + \left( {6{h^2} + 24h} \right){x^2} + \left( {4{h^3} + 24{h^2} + 1} \right)x + \left( {{h^4} + 8{h^3} + h - 5} \right) = 0$ ..... (2)
Now, in order to eliminate the second term from this transformed equation of${x^4} + 8{x^3} + x - 5 = 0$, we have to make the coefficient of \[{x^3}\] zero.
$ \Rightarrow 4h + 8 = 0$
On cross multiplication and simplification we get,
$ \Rightarrow h = \dfrac{{ - 8}}{4} = - 2$
Now, putting h=−2 in (2) we get,
$ \Rightarrow {x^4} + \left( { - 8 + 8} \right){x^3} + \left( {24 - 48} \right){x^2} + \left( { - 32 + 96 + 1} \right)x + \left( {16 - 64 - 2 - 5} \right) = 0$
On simplification we get,
$ \Rightarrow {x^4} - 24{x^2} + 65x - 55 = 0$
Therefore, the transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is ${x^4} - 24{x^2} + 65x - 55 = 0$
Hence, option (A) is correct.
Note: In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. An example of a polynomial of a single indeterminate x is \[{x^2}\; - {\text{ }}4x{\text{ }} + {\text{ }}7\]. Here, we have to eliminate second term so we take coefficient of \[{x^3}\] to be zero, in other questions if it is asked to eliminate any other term we first substitute \[\left( {x + h} \right)\] in place of x, and then take the coefficient of the term which is asked to eliminate, as 0, and then we solve further.
Complete step by step answer:
According to the question we are to eliminate the second term from the equation ${x^4} + 8{x^3} + x - 5 = 0$..... (1)
Let us at first use the substitution \[\left( {x + h} \right)\] in the place of x in equation (1).
Therefore we have,
${\left( {x + h} \right)^4} + 8{\left( {x + h} \right)^3} + \left( {x + h} \right) - 5 = 0$
On expanding we get,
$ \Rightarrow {x^4} + 4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4} + 8\left[ {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right] + x + h - 5 = 0$
On taking terms common we get,
$ \Rightarrow {x^4} + \left( {4h + 8} \right){x^3} + \left( {6{h^2} + 24h} \right){x^2} + \left( {4{h^3} + 24{h^2} + 1} \right)x + \left( {{h^4} + 8{h^3} + h - 5} \right) = 0$ ..... (2)
Now, in order to eliminate the second term from this transformed equation of${x^4} + 8{x^3} + x - 5 = 0$, we have to make the coefficient of \[{x^3}\] zero.
$ \Rightarrow 4h + 8 = 0$
On cross multiplication and simplification we get,
$ \Rightarrow h = \dfrac{{ - 8}}{4} = - 2$
Now, putting h=−2 in (2) we get,
$ \Rightarrow {x^4} + \left( { - 8 + 8} \right){x^3} + \left( {24 - 48} \right){x^2} + \left( { - 32 + 96 + 1} \right)x + \left( {16 - 64 - 2 - 5} \right) = 0$
On simplification we get,
$ \Rightarrow {x^4} - 24{x^2} + 65x - 55 = 0$
Therefore, the transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is ${x^4} - 24{x^2} + 65x - 55 = 0$
Hence, option (A) is correct.
Note: In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. An example of a polynomial of a single indeterminate x is \[{x^2}\; - {\text{ }}4x{\text{ }} + {\text{ }}7\]. Here, we have to eliminate second term so we take coefficient of \[{x^3}\] to be zero, in other questions if it is asked to eliminate any other term we first substitute \[\left( {x + h} \right)\] in place of x, and then take the coefficient of the term which is asked to eliminate, as 0, and then we solve further.
Recently Updated Pages
How do you convert r6sec theta into Cartesian form class 10 maths CBSE

How do you solve dfrac5y3dfracy+72y6+1 and find any class 10 maths CBSE

If sin A+B1 and cos AB1 0circ le left A+B rightle 90circ class 10 maths CBSE

On the number line 10 is to the of zero class 10 maths CBSE

How do you solve 5xge 30 class 10 maths CBSE

In the following sentence supply a verb in agreement class 10 english CBSE

Trending doubts
Write an application to the principal requesting five class 10 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the median of the first 10 natural numbers class 10 maths CBSE

Write examples of herbivores carnivores and omnivo class 10 biology CBSE
