If the sum of the roots of the quadratic equa | vedclass

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

Question

(C) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.Then,$\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.Given tha

Vedclass · Accepted Answer

$H.P.$

Vedclass · Answer

(C) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.Then,$\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.Given that the sum of the roots is equal to the sum of the squares of their reciprocals:$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.Substituting the values:$-b/a = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.$-b/a = \frac{b^2 - 2ac}{c^2} \implies -bc^2 = ab^2 - 2a^2c$.Rearranging gives $2a^2c = ab^2 + bc^2$.Dividing both sides by $abc$,we get $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.This implies that $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$Therefore,their reciprocals $\frac{a}{c}, \frac{b}{a}, \frac{c}{b}$ are in $H.P.$

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is equal to the sum of the squares of their reciprocals,then $a/c, b/a, c/b$ are in

Similar Questions

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

Find the combined product of $(a^{4}+b^{4})(a^{2}+b^{2})(a+b)(a-b)$.

$A$ real root of the equation $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$ is

Solve the given equations and select the correct answer from the given options.
$I. (x+y)^{3} = 1331$
$II. x-y+z = 0$
$III. xy = 28$

Find the value of $x^{3}+y^{3}+z^{3}-3xyz$,when $x+y+z=16$ and $xy+yz+zx=78$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module