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

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(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 $a_1=-8$

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$-9$

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(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 $b=$

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