Multiple choice

lf $\alpha$ and $\beta$ are the roots of $\mathrm{x}^{2}+\mathrm{p}\mathrm{x}+\mathrm{q}=0$ and $\alpha^{4},\ \beta^{4}$ are the roots of $\mathrm{x}^{2}-\mathrm{r}\mathrm{x}+\mathrm{s}=0$, then the equation $\mathrm{x}^{2}-4\mathrm{q}\mathrm{x}+(2\mathrm{q}^{2}-\mathrm{r})=0$ has

  1. two real roots

  2. two negative roots

  3. two positive roots

  4. one positive and one negative root
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D Correct answer
Explanation

Given roots of x^2+px+q=0 are a, b. Then a+b = -p, ab = q. Roots of x^2-rx+s=0 are a^4, b^4. Then a^4+b^4 = r, a^4b^4 = s = q^4. The equation x^2 - 4qx + (2q^2 - r) = 0 has discriminant D = 16q^2 - 4(2q^2 - r) = 8q^2 + 4r. Since r = a^4+b^4 > 0, D > 0. Product of roots is 2q^2 - r = 2q^2 - (a^4+b^4). Since a^4+b^4 > 2(a^2b^2) = 2q^2, the product is negative, implying roots have opposite signs.

AI explanation

Let the roots of x^2 + px + q = 0 be alpha and beta, so their sum is alpha + beta = -p and their product is alpha * beta = q. We are given that alpha^4 and beta^4 are the roots of x^2 - rx + s = 0, making their sum r = alpha^4 + beta^4. This sum can be rewritten using the identity a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2, which factors further as ((a+b)^2 - 2ab)^2 - 2(ab)^2. Substituting the known sum and product gives r = (p^2 - 2q)^2 - 2q^2 = p^4 - 4p^2q + 2q^2. We must analyze the new equation x^2 - 4qx + (2q^2 - r) = 0, and substituting our expression for r makes the constant term 2q^2 - r = -p^4 + 4p^2q. The product of the roots of this new equation is the constant term 1, which is 2q^2 - r, meaning it is negative because r = (p^2 - 2q)^2 - 2q^2 is strictly greater than 2q^2. Since the product of the roots is negative, the equation has one positive and one negative real root.

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