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(A) Let the roots of the given equation $x^4-2 x^3+6 x-21=0$ be $\alpha, \beta, \gamma, \delta$. We want to find the equation whose roots are $\alpha^

Vedclass · Accepted Answer

$x^4-4 x^3-18 x^2-36 x+441=0$

Vedclass · Answer

(A) Let the roots of the given equation $x^4-2 x^3+6 x-21=0$ be $\alpha, \beta, \gamma, \delta$. We want to find the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$. Let $y = x^2$,then $x = \sqrt{y}$. The given equation can be written as $x^4-2 x^3+6 x-21=0$. Rearranging terms: $x^4-21 = 2 x^3-6 x = 2 x(x^2-3)$. Substituting $x^2 = y$: $y^2-21 = 2 x(y-3)$. Squaring both sides: $(y^2-21)^2 = 4 x^2(y-3)^2$. Since $x^2 = y$,we have $(y^2-21)^2 = 4 y(y-3)^2$. Expanding both sides: $y^4-42 y^2+441 = 4 y(y^2-6 y+9)$. $y^4-42 y^2+441 = 4 y^3-24 y^2+36 y$. Rearranging to standard form: $y^4-4 y^3-18 y^2-36 y+441=0$. Replacing $y$ with $x$,the required equation is $x^4-4 x^3-18 x^2-36 x+441=0$.

The equation whose roots are squares of the roots of $x^4-2 x^3+6 x-21=0$ is

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