If a, b, c are in G.P.,then the equations ax^ | vedclass

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

Question

(A) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.The equation $ax^2 + 2bx + c = 0$ can be rewritten as $ax^2 + 2\sqrt{ac}x + c = 0$.This simp

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(A) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.The equation $ax^2 + 2bx + c = 0$ can be rewritten as $ax^2 + 2\sqrt{ac}x + c = 0$.This simplifies to $(\sqrt{a}x + \sqrt{c})^2 = 0$,which gives the repeated root $x = -\sqrt{\frac{c}{a}}$.Since this is a common root,it must satisfy the second equation $dx^2 + 2ex + f = 0$.Substituting $x = -\sqrt{\frac{c}{a}}$ into the second equation:$d(-\sqrt{\frac{c}{a}})^2 + 2e(-\sqrt{\frac{c}{a}}) + f = 0$$d(\frac{c}{a}) - 2e\sqrt{\frac{c}{a}} + f = 0$Dividing the entire equation by $c$:$\frac{d}{a} - 2e\frac{1}{\sqrt{ac}} + \frac{f}{c} = 0$Since $b = \sqrt{ac}$,we have:$\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$.This condition implies that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$

If $a, b, c$ are in $G.P.$,then the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have a common root if $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in

Similar Questions

If one of the roots of the equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is coincident,then the numerical value of $(a + b)$ is

If $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2},$ then the relation between $a, b$ and $c$ is

The condition that one root of the equation $ax^2 + bx + c = 0$ is three times the other is

Solve the given two equations and select the correct option.
$I.$ $x = (1331)^{1/3}$
$II.$ $2y^2 - 17y + 36 = 0$

Find the value of $a^{2}+b^{2}+c^{2}-2ab-2bc-2ca$ when $a=17, b=15$ and $c=13$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module