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

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

Question

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

Vedclass · Accepted Answer

$b/a, -2b/3a$

Vedclass · Answer

(A) The given quadratic equation is $3 a^{2} x^{2}-a b x-2 b^{2}=0$.We can factorize the middle term $-abx$ as $-3abx + 2abx$:$3 a^{2} x^{2}-3 a b x+2 a b x-2 b^{2}=0$Taking common terms from the groups:$3 a x(a x-b)+2 b(a x-b)=0$$(a x-b)(3 a x+2 b)=0$Setting each factor to zero:$a x-b=0 \Rightarrow x=b/a$$3 a x+2 b=0 \Rightarrow x=-2b/3a$Thus,the roots are $b/a$ and $-2b/3a$.

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

Similar Questions

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4\sqrt{ab}x + 1 = 0$ will be

If the product of the roots of the equation $mx^2 + 6x + (2m - 1) = 0$ is $-1$,then the value of $m$ will be:

$I. x^{2}+10 x+24=0$
$II. 4 y^{2}-17 y+18=0$

If $x+y=3$ and $xy=2$,then find the value of $x^3-y^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module