Question

a and b are the roots of 2x^2-3x-6=0. Then the equation whose roots are (1÷a) & (1÷b) is.​

Answer 1

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

Given that,

$\rm \: a \: and \: b \: are \: the \: roots \: of \: {2x}^{2} - 3x - 6 = 0$

We know,

$\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm\implies \:a + b = - \dfrac{( - 3)}{2} = \dfrac{3}{2}$

Also,

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm\implies \:ab = \dfrac{ - 6}{2} = - 3$

Now, Consider Sum of the roots, S

$\rm \: S = \dfrac{1}{a} + \dfrac{1}{b}$

$\rm \: S = \dfrac{b + a}{ab}$

$\rm \: S = \dfrac{a + b}{ab}$

$\rm \: S = \: \dfrac{3}{2} \times \dfrac{ - 1}{3}$

$\rm\implies \:\rm \: S = \: - \: \dfrac{1}{2}$

Now, Product of the roots, P

$\rm \: P = \dfrac{1}{a} \times \dfrac{1}{b}$

$\rm \: P = \dfrac{1}{ab}$

$\rm \: P = \dfrac{1}{ - 3}$

$\rm\implies \:\rm \: P = - \dfrac{1}{3}$

So, the required Quadratic equation is given by

$\rm \: {x}^{2} - Sx + P = 0$

$\rm \: {x}^{2} + \frac{1}{2} x - \frac{1}{3} = 0$

$\rm \: \frac{ {6x}^{2} + 3x - 2}{6} x = 0$

$\rm \: {6x}^{2} + 3x - 2 = 0$

Hence, the quadratic equation is

$\color{green}\rm\implies \:\boxed{ \rm{ \: {6x}^{2} + 3x - 2 = 0 \: }}$

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

Short Cut Trick :-

$\sf \: \alpha \: and \: \beta \: are \: the \: roots \: of \: {ax}^{2} + bx + c = 0, \: then \\ \sf \: the \: quadratic \: equation \: whose \: roots \: are \: \dfrac{1}{ \alpha } \: and \: \dfrac{1}{ \beta } \: is \\ \sf \: \color{green}\boxed{ \rm{ \: {cx}^{2} + bx + a = 0 \: }}$

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

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