Question

Solve question number 160​

Answer 1

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

Given quadratic equation is

$\rm :\longmapsto\: {4x}^{2} + 5k = (5k + 1)x$

can be rewritten as

$\rm :\longmapsto\: {4x}^{2} - (5k + 1)x + 5k = 0$

Let assume that,

$\rm :\longmapsto\: \alpha , \beta \: be \: the \: roots \: of \: {4x}^{2} - (5k + 1)x + 5k = 0$

We know,

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

$\rm \implies\: \alpha + \beta = \dfrac{5k + 1}{4}$

and

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

$\rm \implies\: \alpha \beta = \dfrac{5k}{4}$

Now, Given that

$\rm :\longmapsto\: \alpha - \beta = 1$

$\rm :\longmapsto\: {( \alpha - \beta) }^{2} = 1$

$\rm :\longmapsto\: {\alpha}^{2} + {\beta}^{2} - 2\alpha \: \beta = 1$

can be rewritten as

$\rm :\longmapsto\: {\alpha}^{2} + {\beta}^{2} + 2\alpha \: \beta - 4\alpha \: \beta = 1$

$\rm :\longmapsto\: {(\alpha + \beta)}^{2} - 4\alpha \: \beta = 1$

On substituting the values, we get

$\rm :\longmapsto\: {\bigg[\dfrac{5k + 1}{4} \bigg]}^{2} - 4 \times \dfrac{5k}{4} = 1$

$\rm :\longmapsto\:\dfrac{ {(5k + 1)}^{2} }{16} - 5k = 1$

$\rm :\longmapsto\:\dfrac{ 25 {k}^{2} + 1 + 10k}{16}= 5k + 1$

$\rm :\longmapsto\:25 {k}^{2} + 1 + 10k= 80k + 16$

$\rm :\longmapsto\:25 {k}^{2} - 70k - 15 = 0$

$\rm :\longmapsto\:5 {k}^{2} - 14k - 3= 0$

$\rm :\longmapsto\:5 {k}^{2} - 15k + k - 3= 0$

$\rm :\longmapsto\:5k(k - 3) + 1(k - 3) = 0$

$\rm :\longmapsto\:(5k + 1)(k - 3) = 0$

$\bf\implies \:k = - \: \dfrac{1}{5} \: \: or \: \: k = 3$

• Hence, option (c) is Correct.

More to know :-

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