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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

If the roots of the equation $bx^2 + cx + a = 0$ are imaginary,then for all real values of $x,$ the expression $3b^2x^2 + 6bcx + 2c^2$ is :

Let $a, b$ and $c$ be three positive real numbers such that the sum of any two of them is greater than the third. All the values of $\lambda$ such that the roots of the equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ are real,are given by

Let $f(x)=ax^{2}+bx+c$ and $g(x)=px^{2}+qx+r$ such that $f(1)=g(1)$, $f(2)=g(2)$ and $f(3)-g(3)=2$. Then, $f(4)-g(4)$ is:

If the quadratic equation $(\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0, \lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:

The equation $(\cos p - 1) x^2 + (\cos p) x + \sin p = 0$ in the variable $x$ has real roots. Then $p$ can take any value in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module