The roots of 4x^2 + 6px + 1 = 0 are equal,the | vedclass

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

Question

(C) For a quadratic equation $ax^2 + bx + c = 0$,the roots are equal if the discriminant $D = b^2 - 4ac = 0$.Given equation: $4x^2 + 6px + 1 = 0$.Here

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) For a quadratic equation $ax^2 + bx + c = 0$,the roots are equal if the discriminant $D = b^2 - 4ac = 0$.Given equation: $4x^2 + 6px + 1 = 0$.Here,$a = 4$,$b = 6p$,and $c = 1$.Substituting these values into the discriminant formula:$(6p)^2 - 4(4)(1) = 0$$36p^2 - 16 = 0$$36p^2 = 16$$p^2 = \frac{16}{36} = \frac{4}{9}$$p = \pm \sqrt{\frac{4}{9}} = \pm \frac{2}{3}$.Since the options provided only include the positive value,the correct option is $\frac{2}{3}$.

The roots of $4x^2 + 6px + 1 = 0$ are equal,then the value of $p$ is

Similar Questions

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation $(-2+\sqrt{3})(|\sqrt{x}-3|) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0$,$x \ge 0$,then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

Let $\alpha, \alpha+2 \in Z$ be the roots of the quadratic equation $x(x+2) + (x+1)(x+3) + (x+2)(x+4) + \dots + (x+n-1)(x+n+1) = 4n$ for some $n \in N$. Then $n+\alpha$ is equal to:

Solve the equation $\sqrt{2} x^{2} + x + \sqrt{2} = 0$.

If the roots of the equation $x^2 - (3k - 1)x + 2k^2 + 2k = 0$ are equal,then the value of $k$ will be .....

Solve the equation,$3^{x^2-x}=25-4^{x^2-x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module