
The transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is.
$\eqalign{
& A){x^4} - 24{x^2} + 65x - 55 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \cr
& B){x^4} + 24{x^2} + 65x + 55 = 0\, \cr
& C){x^4} - 24{x^2} - 65x + 55 = 0\, \cr
& D){x^4} + 24{x^2} + 65x - 55 = 0\, \cr} $.
Answer
586.5k+ views
Hint:
We have to transform the given equation in x to the equation in y, substituting x by (y+h). Then equating the coefficient of the term containing power of y as 3 to zero, we get the required equation in y as well as x(by putting y=x).
Complete step by step solution:
Step1: The given equation is ${x^4} + 8{x^3} + x - 5 = 0$ …………(1)
Step2: Substituting x by (y+h) in equation (1).
Then the equation becomes,
$\eqalign{
& {(y + h)^4} + 8{(y + h)^3} + (y + h) - 5 = 0 \cr
& or,{y^4} + {}^4{C_1}{y^3}h + {}^4{C_2}{y^2}{h^2} + {}^4{C_3}y{h^3} + {h^4} + 8({y^3} + 3{y^2}h + 3y{h^2} + {h^3}) + y + h - 5 = 0 \cr
& or,{y^4} + 4{y^3}h + 6{y^2}{h^2} + 4y{h^3} + {h^4} + 8{y^3} + 24{y^2}h + 24y{h^2} + 8{h^3} + y + h - 5 = 0 \cr
& or,{y^4} + (4h + 8){y^3} + (6{h^2} + 24h){y^2} + (4{h^3} + 24{h^2} + 1)y + {h^4} + 8{h^3} + h - 5 = 0.............(2) \cr} $
Step3: Here the coefficient of ${y^3}$ in second term in equation (2) is $(4h + 8)$ .
Step4: To vanish the second term,
$\eqalign{
& 4h + 8 = 0 \cr
& or,h = - 2 \cr} $.
Step5: Putting the value of h in (2), we get
$\eqalign{
& {y^4} + \{ 6 \times 4 + 24 \times ( - 2)\} {y^2} + \{ 4 \times ( - 8) + 24 \times 4 + 1\} y + 16 - 64 - 2 - 5 = 0 \cr
& or,{y^4} - 24{y^2} + 65y - 55 = 0\,\,\,\,\,\,\,\,\,\,\,..............(3) \cr} $.
This is the required equation in ‘y’.
Step6: Substituting y by x in the last equation (3), we get
${x^4} - 24{x^2} + 65x - 55 = 0$
This is the required equation.Hence, the option A) is correct here.
Note:
Here we have eliminated the second term where the power of the variable is 3. In a similar way,we can eliminate any term by judging the value of h there. This is called origin shifting. We’ll shift the origin to that point which we need to illuminate.
We have to transform the given equation in x to the equation in y, substituting x by (y+h). Then equating the coefficient of the term containing power of y as 3 to zero, we get the required equation in y as well as x(by putting y=x).
Complete step by step solution:
Step1: The given equation is ${x^4} + 8{x^3} + x - 5 = 0$ …………(1)
Step2: Substituting x by (y+h) in equation (1).
Then the equation becomes,
$\eqalign{
& {(y + h)^4} + 8{(y + h)^3} + (y + h) - 5 = 0 \cr
& or,{y^4} + {}^4{C_1}{y^3}h + {}^4{C_2}{y^2}{h^2} + {}^4{C_3}y{h^3} + {h^4} + 8({y^3} + 3{y^2}h + 3y{h^2} + {h^3}) + y + h - 5 = 0 \cr
& or,{y^4} + 4{y^3}h + 6{y^2}{h^2} + 4y{h^3} + {h^4} + 8{y^3} + 24{y^2}h + 24y{h^2} + 8{h^3} + y + h - 5 = 0 \cr
& or,{y^4} + (4h + 8){y^3} + (6{h^2} + 24h){y^2} + (4{h^3} + 24{h^2} + 1)y + {h^4} + 8{h^3} + h - 5 = 0.............(2) \cr} $
Step3: Here the coefficient of ${y^3}$ in second term in equation (2) is $(4h + 8)$ .
Step4: To vanish the second term,
$\eqalign{
& 4h + 8 = 0 \cr
& or,h = - 2 \cr} $.
Step5: Putting the value of h in (2), we get
$\eqalign{
& {y^4} + \{ 6 \times 4 + 24 \times ( - 2)\} {y^2} + \{ 4 \times ( - 8) + 24 \times 4 + 1\} y + 16 - 64 - 2 - 5 = 0 \cr
& or,{y^4} - 24{y^2} + 65y - 55 = 0\,\,\,\,\,\,\,\,\,\,\,..............(3) \cr} $.
This is the required equation in ‘y’.
Step6: Substituting y by x in the last equation (3), we get
${x^4} - 24{x^2} + 65x - 55 = 0$
This is the required equation.Hence, the option A) is correct here.
Note:
Here we have eliminated the second term where the power of the variable is 3. In a similar way,we can eliminate any term by judging the value of h there. This is called origin shifting. We’ll shift the origin to that point which we need to illuminate.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

