Multiple choice

$\alpha$ and $\beta$ are the roots of the equation $ax^{2}+bx+c=0$ and $\alpha^{4}, \beta^{4}$ are the roots of the equation $lx^{2} +mx+n=0(\alpha, \beta$ are real and distinct.) Let $f(x)=a^{2}lx^{2}-4aclx+2c^{2}l+a^{2}m=0$, then Roots of $f(x)=0$ are

  1. real and same in sign

  2. real and opposite in sign

  3. equal

  4. data is insufficient

Reveal answer Fill a bubble to check yourself
B Correct answer
AI explanation

Using the relationships between roots and coefficients, we express the sum and product of the new roots as a^4 + b^4 = -(m/l) and a^4 * b^4 = n/l. The discriminant of the given quadratic f(x) = (al)^2 x^2 - 4acl x + (2c^2 l + a^2 m) is evaluated as D = (-4acl)^2 - 4(al)^2(2c^2 l + a^2 m). Simplifying this yields 16a^2c^2l^2 - 8a^2c^2l^2 - 4a^4l^2m, which reduces to 8a^2c^2l^2 - 4a^4l^2m. Substituting m = -l(a^4 + b^4) into the expression gives D = 4a^2l^2(2c^2 + a^2(a^4 + b^4)). Since a^4 + b^4 is always positive for real, distinct roots, the discriminant is strictly positive, meaning the roots are real. The product of the roots of f(x) is (2c^2 l + a^2 m) divided by (a^2 l), which simplifies to 2c^2 - a^2(a^4 + b^4) using the substitution for m. Expanding and rearranging the term a^2(a^4 + b^4) reveals it equals 2a^2b^2, making the product 2c^2 - 2a^2b^2 = 2(c^2 - a^2b^2), which must be negative because the original roots are real and distinct, ensuring c^2 < a^2b^2. A positive discriminant and a negative product confirm the roots are real and opposite in sign. The correct result is real and opposite in sign.