If the absolute difference between the roots | vedclass

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

Question

(B) Let the roots of the equation $x^2 + px + 3 = 0$ be $\alpha$ and $\beta$.Given that $|\alpha - \beta| = \sqrt{p}$.Squaring both sides,we get $(\al

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let the roots of the equation $x^2 + px + 3 = 0$ be $\alpha$ and $\beta$.Given that $|\alpha - \beta| = \sqrt{p}$.Squaring both sides,we get $(\alpha - \beta)^2 = p$.We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.Substituting the coefficients from the quadratic equation,$\alpha + \beta = -p$ and $\alpha\beta = 3$.So,$(-p)^2 - 4(3) = p$.$p^2 - 12 = p \implies p^2 - p - 12 = 0$.Factoring the quadratic equation: $(p - 4)(p + 3) = 0$.Thus,$p = 4$ or $p = -3$.Since the difference between the roots is $\sqrt{p}$,$p$ must be non-negative for the roots to be real and the expression to be defined in the real number system. Therefore,$p = 4$.

If the absolute difference between the roots of the equation $x^2 + px + 3 = 0$ is $\sqrt{p}$,then $p = ......$

Similar Questions

If for any real $x$,$y = \frac{11 x^2+12 x+6}{x^2+4 x+2}$ is such that $y < a$ or $y \geq b$,then $a, b$ are

The value of $k$ for which the quadratic equation $kx^2 + 1 = kx + 3x - 11x^2$ has real and equal roots is:

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-Px^2+Qx-R=0$ and $(\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2$ are the roots of the equation $x^3-5x^2+4x=0$,then the possible least value of $P+Q+R$ is

If the equation $(m - n)x^2 + (n - l)x + l - m = 0$ has equal roots,then $l, m,$ and $n$ satisfy:

Let $a \ne a_1 \ne 0,$ $f(x) = ax^2 + bx + c,$ $g(x) = a_1x^2 + b_1x + c_1,$ and $p(x) = f(x) - g(x).$ If $p(x) = 0$ only for $x = -1$ and $p(-2) = 2,$ then the value of $p(2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module