Question

Solve the quadratic equation $3{a^2}{x^2} + 8abx + 4{b^2} = 0,{\text{ a}} \ne {\text{0}}$
A. No, Roots are imaginary
B. $\dfrac{{ - 2b}}{a}, - \dfrac{{2b}}{{3a}}$, real
C. Cannot be determined
D. incorrect question

Answer 1

The basic idea in this question is that the given equation is in the form of quadratic equation whose general form is given by $A{x^2} + Bx + C$ , where A, B and C are constants and their values can be determined by comparing it with the given quadratic equation.

Complete step-by-step solution:
The given equation is,
$3{a^2}{x^2} + 8abx + 4{b^2} = 0$
$\because$ It is a quadratic equation whose general equation is $A{x^2} + Bx + C$ ,
$\therefore$ On comparing,
$A = 3{a^2}$
$B = 8ab$
$C = 4{b^2}$
$\because$ It is a quadratic equation it can be solved by applying quadratic formula given by $x = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
On putting the values of A, B and C,
$\Rightarrow x = \dfrac{{ - 8ab \pm \sqrt {{{\left( {8ab} \right)}^2} - 4 \times 3{a^2} \times 4{b^2}} }}{{2 \times 3{a^2}}}$
On simplifying further,
$\Rightarrow x = \dfrac{{ - 8ab \pm \sqrt {64{a^2}{b^2} - 48{a^2}{b^2}} }}{{6{a^2}}}$
On subtracting the terms inside square brackets,
$\Rightarrow x = \dfrac{{ - 8ab \pm \sqrt {16{a^2}{b^2}} }}{{6{a^2}}}$
$\Rightarrow x = \dfrac{{ - 8ab \pm \sqrt {{{\left( {4ab} \right)}^2}} }}{{6{a^2}}}$
On solving the square root,
$\Rightarrow x = \dfrac{{ - 8ab \pm 4ab}}{{6{a^2}}}$
$\Rightarrow x = \dfrac{{ - 8ab + 4ab}}{{6{a^2}}}$
And $x = \dfrac{{ - 8ab - 4ab}}{{6{a^2}}}$
When $x = \dfrac{{ - 8ab + 4ab}}{{6{a^2}}}$
On simplifying,
$\Rightarrow x = \dfrac{{ - 4ab}}{{6{a^2}}}$
On dividing numerator and denominator by 2a in RHS,
$\Rightarrow x = - \dfrac{{2b}}{{3a}}$
When $x = \dfrac{{ - 8ab - 4ab}}{{6{a^2}}}$
On simplifying,
$x = \dfrac{{ - 12ab}}{{6{a^2}}}$
On dividing numerator and denominator by 6a in RHS,
$\Rightarrow x = \dfrac{{ - 2b}}{a}$
$\therefore$ The given quadratic equation has two solutions,
$x = - \dfrac{{2b}}{{3a}}$
And,
$x = \dfrac{{ - 2b}}{a}$
$\therefore$ Option (B) is correct.

Note: In this type of question the equation can be solved directly with the quadratic formula and hence the formula must be known. Calculations should be done carefully to avoid any mistake. In the final answer the values of the constants may or may not be given. If the value of the discriminant i.e. $\sqrt {{B^2} - 4AC}$ comes negative then the equation has imaginary roots. If the value of the discriminant is zero then the equation has equal roots and if the value of the discriminant is positive then the equation has real roots as in this question. Always try to solve step by step. After the final answer is found out it can be checked whether it satisfies the original equation given in the question by simply substituting its value in the equation.