Roots of ax^2 + b = 0 are real and distinct i | vedclass

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

Question

(B) For a quadratic equation of the form $Ax^2 + Bx + C = 0$,the roots are real and distinct if the discriminant $D = B^2 - 4AC > 0$.In the given equa

Vedclass · Accepted Answer

$ab < 0$

Vedclass · Answer

(B) For a quadratic equation of the form $Ax^2 + Bx + C = 0$,the roots are real and distinct if the discriminant $D = B^2 - 4AC > 0$.In the given equation $ax^2 + 0x + b = 0$,we have $A = a$,$B = 0$,and $C = b$.Substituting these values into the discriminant formula:$D = (0)^2 - 4(a)(b) > 0$$-4ab > 0$Dividing both sides by $-4$ (which reverses the inequality sign):$ab Therefore,the roots are real and distinct if $ab < 0$.

Roots of $ax^2 + b = 0$ are real and distinct if

Similar Questions

If $\alpha$ and $\beta$ are the roots of $x^2 - 3x + a = 0$,where $a \in R$ and $\alpha < 1 < \beta$,then:

$I. 3x^2 - 20x - 32 = 0$
$II. 2y^2 - 3y - 20 = 0$

If $x$ is a rational number and $\frac{(x+1)^{3}-(x-1)^{3}}{(x+1)^{2}-(x-1)^{2}}=2$,then the sum of the numerator and denominator of $x$ is

All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot \frac{1}{4^{\sin^2 y}} \leq 1$ also satisfy the equation:

If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$,then the value of $\alpha^n + \beta^n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module