Class 11Mathematics4-2.Quadratic Equations and InequationsSolution of quadratic equations and Nature of roots
MediumThe quadratic equation $2x^{2} - (a^{3} + 8a - 1)x + a^{2} - 4a = 0$ has roots of opposite signs. Then,
- A$a \leq 0$
- B$0 < a < 4$
- C$4 \leq a < 8$
- D$a \geq 8$
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Similar Questions
What is the product of the real roots of the equation ${t^2}{x^2} + |x| + 9 = 0$?
The equation $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$ has the solution:
Difficult
View SolutionWith respect to the roots of the equation $3x^3 + bx^2 + bx + 3 = 0$,match the items of List-$I$ with those of List-$II$.
List-$I$ List-$II$ $A$. All the roots are negative $I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$ $B$. Two roots are complex $II$. $-3 < b < 9$ $C$. Two roots are positive $III$. $b \in (-\infty, -3) \cup (9, \infty)$ $D$. All roots are real and distinct $IV$. $b = 9$ $V$. $b = -3$
| List-$I$ | List-$II$ |
| $A$. All the roots are negative | $I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$ |
| $B$. Two roots are complex | $II$. $-3 < b < 9$ |
| $C$. Two roots are positive | $III$. $b \in (-\infty, -3) \cup (9, \infty)$ |
| $D$. All roots are real and distinct | $IV$. $b = 9$ |
| $V$. $b = -3$ |
$\cos(2x + 7) = a(2 - \sin x)$ can have a real solution for
DifficultWBJEE 2020
View SolutionIf $\sin A, \sin B, \cos A$ are in $G.P.$,then the roots of $x^2 + 2x \cot B + 1 = 0$ are always ......
Medium
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