Question

find the roots of the quadratic equation 15a^2x^2-16abx-15b^2=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm \: 15 {a}^{2} {x}^{2} - 16abx - 15 {b}^{2} = 0$

On splitting the middle terms, we get

$\rm \: 15 {a}^{2} {x}^{2} - 25abx + 9abx - 15 {b}^{2} = 0$

$\rm \: 5ax(3ax - 5b) + 3b(3ax - 5b) = 0$

$\rm \: (3ax - 5b)(5ax + 3b) = 0$

$\rm \: 3ax - 5b = 0 \: \: or \: \: 5ax + 3b = 0$

$\rm \: 3ax = 5b\: \: or \: \: 5ax = - 3b$

$\bf\implies \:x = \dfrac{5b}{3a} \: \: or \: \: x = \: - \: \dfrac{3b}{5a}$

$\rule{190pt}{2pt}$

Concept Used :-

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers m and n such that m + n = b and mn = ac.

After finding m and n, we split the middle term i.e bx in the quadratic equation as mx + nx and get the required factors by grouping the terms.

$\rule{190pt}{2pt}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

Answer 2

$\huge{ \colorbox{black}{ \color{lime}{Añswèr}}}$

$\large{ \sf \underline{Solution:-}}$

Given quadratic equation is

$\sf \: 15 {a}^{2} {x}^{2} - 16abx - 15 {b}^{2} = 0$

On splitting the middle terms, we get

$\sf \: {15a}^{2} {x}^{2} - 25abx + 9abx - 15 {b}^{2} = 0$

$\sf \: 5ax(3ax - 5b) + 3b(3ax - 5b) = 0$

$\sf \: (3ax - 5b)(5ax + 3b) = 0$

$\sf \: \: 3ax - 5b = 0 \: or \: 5ax + 3b = 0$

$\sf \: \: 3ax = 5b \: or \: 5ax = - 3b$

$\longmapsto \bf \: x = \frac{5b}{3a} \: \: \sf \: or \: x = - \frac{3b}{5a}$

$\rule{1333pt}{1.8pt}$

$\bf \pmb {\frak{be \: brainly..}}$