Question

Find the algebraic equation whose roots are two times of its roots x^5-2x^+3x^3-2x^2+4x+3

Answer 1

$\textbf{Given:}$

$\textsf{Equation is}$

$\mathsf{x^5-2x^4+3x^3-2x^2+4x+3=0}$

$\textbf{To find:}$

$\textsf{The equation whose roots are two times of its roots of}\;\mathsf{x^5-2x^4+3x^3-2x^2+4x+3=0}$

$\textbf{Solution:}$

$\mathsf{Let\;\alpha\;be\;a\;root\;of\;the\;given\;equation}$

$\mathsf{Then,\;\alpha^5-2\alpha^4+3\alpha^3-2\alpha^2+4\alpha+3=0}$.........(1)

$\mathsf{Take\;2\alpha=x\,\implies\,\alpha=\dfrac{x}{2}}$

$\textsf{Now, (1) becomes}$

$\mathsf{\left(\dfrac{x}{2}\right)^5-2\left(\dfrac{x}{2}\right)^4+3\left(\dfrac{x}{2}\right)^3-2\left(\dfrac{x}{2}\right)^2+4\left(\dfrac{x}{2}\right)+3=0}$

$\mathsf{\dfrac{x^5}{32}-2\left(\dfrac{x^4}{16}\right)+3\left(\dfrac{x^3}{8}\right)-2\left(\dfrac{x^2}{4}\right)+4\left(\dfrac{x}{2}\right)+3=0}$

$\mathsf{\dfrac{x^5-4x^4+12x^3-16x^2+64x+96}{32}=0}$

$\boxed{\mathsf{x^5-4x^4+12x^3-16x^2+64x+96=0}}$

$\textsf{which is the required equation whose roots are}$

$\textsf{two times the roots of the given equatioon}$

$\textbf{Find more:}$

If two of the roots of equation x4 - 2x3 + ax2 + 8x + b = 0 are equal in magnitude but opposite in sig

then value of 4a + b is equal to :

(A) 16

(B) 8

(C) -16

theme values are also

28 ** ** *5 = o are equal in magnitude but opposite in

(D) -8​

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