Let the transformed equation of 2x^4-8x^3+3x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation $2x^4-8x^3+3x^2-1=0$. To remove the $x^3$ term,we use the transformation $x = y - \frac{a_1}{n a_0}$,where $a_0=2$ and $a_1=-8$

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

(C) Given the equation $2x^4-8x^3+3x^2-1=0$. To remove the $x^3$ term,we use the transformation $x = y - \frac{a_1}{n a_0}$,where $a_0=2$ and $a_1=-8$ for a quartic equation $a_0x^4+a_1x^3+a_2x^2+a_3x+a_4=0$. Here,$h = -\frac{-8}{4 	imes 2} = \frac{8}{8} = 1$. Substitute $x = y+1$ into the equation: $2(y+1)^4 - 8(y+1)^3 + 3(y+1)^2 - 1 = 0$. Expanding the terms: $2(y^4+4y^3+6y^2+4y+1) - 8(y^3+3y^2+3y+1) + 3(y^2+2y+1) - 1 = 0$. $2y^4 + 8y^3 + 12y^2 + 8y + 2 - 8y^3 - 24y^2 - 24y - 8 + 3y^2 + 6y + 3 - 1 = 0$. Combining like terms: $2y^4 + (8-8)y^3 + (12-24+3)y^2 + (8-24+6)y + (2-8+3-1) = 0$. $2y^4 - 9y^2 - 10y - 4 = 0$. Comparing this with $2x^4+bx^2+cx+d=0$,we find $b = -9$.

Let the transformed equation of $2x^4-8x^3+3x^2-1=0$ such that the term containing the cubic power of $x$ is absent be $2x^4+bx^2+cx+d=0$. Then $b=$

Similar Questions

If $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$ for all $x \in R$,then the interval of maximum length in which $y$ lies is

Let $r$ be a root of the equation $x^2+2x+6=0$. The value of $(r+2)(r+3)(r+4)(r+5)$ is equal to

The least value of $\frac{x^2y^2 - 2x^2y + 2x^2 + 2xy - 2x + 1}{x^2y + x}$ is $\lambda$,where $x, y \in R^+$ and $x^2y + x \neq 0$. Then:

If $a, b, c \in \mathbb{R}$ and $1$ is a root of the equation $ax^2 + bx + c = 0$,then the curve $y = 4ax^2 + 3bx + 2c$ $(a \neq 0)$ intersects the $x$-axis at

The sum of all integral values of $k$ $(k \neq 0)$ for which the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ in $x$ has no real roots,is ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module