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Question

if $$\alpha$$ and $$\beta$$ are roots of $$x^{2}+px+q=0$$ and $$\alpha^{4},\beta^{4}$$ are roots of $$x^{2}-rx+s=0$$, then prove that the equation $$x^{2}-4qx+2q^{2}-r=0$$ has distinct and real roots.


Solution

The equation $$x^2-4qx+2q^2-r=0$$
$$D=b^2-4ac$$
$$\therefore D=(-42)^2-4(1)(2q^2-r)$$
$$=16q^2-8q^2+4r$$ $$=\sqrt{8q^2}+4r$$
Now $$\alpha^4+\beta^4=\dfrac{-(-r)}{1}=r$$
$$\therefore \alpha^4+\beta^4 > 0$$ $$(\alpha_1\beta \neq 0)$$
$$\therefore r > 0$$
$$\therefore 8q^2+4r > 0$$ $$(D > 0)$$ proved
$$\therefore$$ The equation $$x^2-4qx+2q^2-r=0$$ has real and distinct roots.

Mathematics

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