Let a, b, c be real numbers with a ne 0. If a | vedclass

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

Question

(D) Given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$,we have $a^2\alpha^2 + b\alpha + c = 0$,which implies $b\alpha + c = -a^2\alpha^2$.Given th

Vedclass · Accepted Answer

$\alpha < \gamma < \beta$

Vedclass · Answer

(D) Given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$,we have $a^2\alpha^2 + b\alpha + c = 0$,which implies $b\alpha + c = -a^2\alpha^2$.Given that $\beta$ is a root of $a^2x^2 - bx - c = 0$,we have $a^2\beta^2 - b\beta - c = 0$,which implies $b\beta + c = a^2\beta^2$.Let $f(x) = a^2x^2 + 2bx + 2c$. We evaluate $f(x)$ at $\alpha$ and $\beta$:$f(\alpha) = a^2\alpha^2 + 2(b\alpha + c) = a^2\alpha^2 + 2(-a^2\alpha^2) = -a^2\alpha^2$. Since $a e 0$ and $\alpha > 0$,$f(\alpha) $f(\beta) = a^2\beta^2 + 2(b\beta + c) = a^2\beta^2 + 2(a^2\beta^2) = 3a^2\beta^2$. Since $a e 0$ and $\beta > 0$,$f(\beta) > 0$.Since $f(\alpha) 0$,by the Intermediate Value Theorem,there must exist a root $\gamma$ of $f(x) = 0$ such that $\alpha < \gamma < \beta$.

Let $a, b, c$ be real numbers with $a \ne 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$,$\beta$ is a root of $a^2x^2 - bx - c = 0$,and $0 < \alpha < \beta$,then the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies:

Similar Questions

If $(x-2)(x-p) = x^{2}-ax+6$,then the value of $(a-p)$ is

The number of solutions of $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $y^{2}-9y+20=0$

If $\lambda$ is the ratio of the roots of the quadratic equation in $x$,$3m^2x^2 + m(m - 4)x + 2 = 0$,then the least value of $m$ for which $\lambda + \frac{1}{\lambda} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module