For which real value of a do the roots of the | vedclass

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(B) For the roots of a quadratic equation $Ax^2 + Bx + C = 0$ to have opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} < 0$.

Vedclass · Accepted Answer

$0 < a < 4$

Vedclass · Answer

(B) For the roots of a quadratic equation $Ax^2 + Bx + C = 0$ to have opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} Given the equation $2x^2 - (a^3 + 8a - 1) x + (a^2 - 4a) = 0$,we have $A = 2$ and $C = a^2 - 4a$.The condition for opposite signs is $\frac{a^2 - 4a}{2} This implies $a^2 - 4a Factoring the expression,we get $a(a - 4) This inequality holds when $0 Since the discriminant $D = B^2 - 4AC$ is always non-negative for these values of $a$ (as $AC  0$),the roots are guaranteed to be real and distinct.Thus,the correct range is $0 < a < 4$.

For which real value of $a$ do the roots of the quadratic equation $2x^2 - (a^3 + 8a - 1) x + a^2 - 4a = 0$ have opposite signs?

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