Question

Solve 3a²x² + 11abx + 6b² = 0 , a ≠ 0,
Find out the solution set,
Ans : { -3b/a , -2b/3a }

Answer 1

Answer:

3a^2x^2 + 11abx + 6b^2 = 0

=> 3a^2x^2 + 9abx + 2abx + 6 b^2 = 0

=> 3ax (ax + 3b) + 2b (ax + 3b) = 0

=> (ax + 3b) (3ax + 2b) = 0 [ By Factorization ]

=> ax + 3b = 0 or 3ax + 2b = 0

[ Zero product rule ]

Step-by-step explanation:

@darkaoul

Answer 2

Given :

$3a^2x^2+11abx+6b^2=0 \: ; \: a\neq0$

To find :

We have to find the roots/solutions for the given equation/polynomial.

Solution :

The polynomial $3a^2x^2+11abx+6b^2=0$ where $a\neq0$, is similar to that of a quadratic equation which is in the form of $ax^2+bx+c=0~~[a\neq0]$. Here :

• $a = 3a^2$

• $b = 11ab$

• $c = 6b^2$

Now, by the quadratic formula, we'll find the roots/solutions :

• First root/solution :

${\underline {\boxed{{x= \frac{-b + \sqrt{b^2-4ac} }{2a } }}}}$

$:\implies x= \dfrac{-(11ab) + \sqrt{(11ab)^2-4(3 {a}^{2} )(6 {b}^{2} )} }{2(3 {a}^{2} )}$

$:\implies x= \dfrac{-11ab+ \sqrt{121a^2b^2-4(18 {a}^{2} {b}^{2} )} }{6 {a}^{2} }$

$:\implies x= \dfrac{-11ab + \sqrt{121 {a}^{2} {b}^{2} - 72 {a}^{2} {b}^{2} } }{6 {a}^{2} }$

$:\implies x= \dfrac{-11ab + \sqrt{49a^2b^2} }{6a^2}$

$:\implies x= \dfrac{ - 11ab + 7ab}{6 {a}^{2} }$

$:\implies x= \dfrac{ - 4ab}{6 {a}^{2} }$

$:\implies x= \dfrac{-\cancel{2a}(2b)}{\cancel{2a}(3a)}$

$\underline {\boxed{\therefore \: x= \frac{-2b}{3a} }}$

• Second root/solution :

$\underline{\boxed{x= \dfrac{-b- \sqrt{b^2-4ac}}{2a}}}$

$:\implies x= \dfrac{-11ab - \sqrt{(11ab)^2-4(3a^2)(6b^2)} }{2(3a)^2}$

$:\implies x= \dfrac{ - 11ab \sqrt{121 {a}^{2} {b}^{2} - 4(18 {a}^{2} {b}^{2}) } }{6 {a}^{2} }$

$:\implies x= \dfrac{-11ab- \sqrt{121a^2b^2 - 72a^2b^2} }{6 {a}^{2} }$

$:\implies x= \dfrac{ - 11ab - \sqrt{49 {a}^{2} {b}^{2} } }{6 {a}^{2} }$

$:\implies x= \dfrac{ - 11ab - 7ab}{6 {a}^{2} }$

$:\implies x= \dfrac{ - 18ab}{6 {a}^{2} }$

$:\implies x= \dfrac{ - \cancel{6a}(3b)}{\cancel{6a}(a)}$

$\underline {\boxed{\therefore \: x= \dfrac{-3b}{a} }}$

_____________________

Hence, the roots are $\dfrac{-2b}{3a}$ and $\dfrac{-3b}{a}$.