The roots of the equation a^{2} x^{2}-3 a b x | vedclass

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

Question

(B) The given quadratic equation is $a^{2} x^{2}-3 a b x+2 b^{2}=0$.We can factorize the quadratic expression by splitting the middle term:$a^{2} x^{2

Vedclass · Accepted Answer

$\frac{2 b}{a}, \frac{b}{a}$

Vedclass · Answer

(B) The given quadratic equation is $a^{2} x^{2}-3 a b x+2 b^{2}=0$.We can factorize the quadratic expression by splitting the middle term:$a^{2} x^{2}-2 a b x-a b x+2 b^{2}=0$Grouping the terms,we get:$a x(a x-2 b)-b(a x-2 b)=0$Taking $(a x-2 b)$ as a common factor:$(a x-2 b)(a x-b)=0$Setting each factor to zero:$a x-2 b=0 \Rightarrow x=\frac{2 b}{a}$$a x-b=0 \Rightarrow x=\frac{b}{a}$Thus,the roots of the equation are $\frac{2 b}{a}$ and $\frac{b}{a}$.

The roots of the equation $a^{2} x^{2}-3 a b x+2 b^{2}=0$ are

Similar Questions

Let $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4 + x^2 + 1 = 0$. Then,the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$ is:

Two positive distinct numbers $a$ and $b$ each differ from their reciprocal by $1$. The value of $a + b$ is

The sum of the solutions of the equation $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$ $(x > 0)$ is equal to

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+12x+36=0$
$II.$ $y^{2}+15y+56=0$

If the roots of the equation $qx^2 + px + q = 0$,where $p$ and $q$ are real,are complex,then the roots of the equation $x^2 - 4qx + p^2 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module