Question

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 1

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.