Question

if one zero of quadratic polynomial x square -x -k is the square of other then find the k​

Answer 1

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

Let assume that

$\rm :\longmapsto\: \alpha , { \alpha }^{2} \: are \: zeroes \: of \: {x}^{2} - x - k$

We know that,

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

$\bf\implies \:\alpha \times {\alpha}^{2} = \dfrac{ - k}{1}$

$\bf\implies \: {\alpha}^{3} \: = - \: k$

Also,

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

$\bf\implies \:\alpha + {\alpha}^{2} = - \dfrac{( - 1)}{1} = 1$

On cubing both sides, we get

$\rm :\longmapsto\: {(\alpha + {\alpha}^{2} )}^{3} = {1}^{3}$

We know,

$\red{\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}}$

$\rm :\longmapsto\: {\alpha}^{3} + {\alpha}^{6} + 3 \times \alpha \times {\alpha}^{2}(\alpha + {\alpha}^{2}) = 1$

As,

$\boxed{ \bf{ \: {\alpha}^{3} = - k \: \: \: and \: \: \: \alpha + {\alpha}^{2} = 1}}$

And on substituting these values, we ge

ge$\rm :\longmapsto\: - k + {( - k)}^{2} + 3( - k)(1) = 1$

ge$\rm :\longmapsto\: - k + {( - k)}^{2} + 3( - k)(1) = 1$$\rm :\longmapsto\: - k + {k}^{2} - 3k - 1 = 0$

ge$\rm :\longmapsto\: - k + {( - k)}^{2} + 3( - k)(1) = 1$$\rm :\longmapsto\: - k + {k}^{2} - 3k - 1 = 0$$\rm :\longmapsto\: {k}^{2} - 4k - 1 = 0$

ge$\rm :\longmapsto\: - k + {( - k)}^{2} + 3( - k)(1) = 1$$\rm :\longmapsto\: - k + {k}^{2} - 3k - 1 = 0$$\rm :\longmapsto\: {k}^{2} - 4k - 1 = 0$Its a quadratic in k, so to get the value of k, we use Quadratic Formula,

$\rm :\longmapsto\:k = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

Here,

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = - 4$

$\rm :\longmapsto\:c = - 1$

So, on substituting the values of a, b, c, we get

$\rm :\longmapsto\:k = \dfrac{ - ( - 4) \: \pm \: \sqrt{ {( - 4)}^{2} - 4(1)( - 1) } }{2(1)}$

$\rm :\longmapsto\:k = \dfrac{4 \: \pm \: \sqrt{16 + 4 } }{2}$

$\rm :\longmapsto\:k = \dfrac{4 \: \pm \: \sqrt{20} }{2}$

$\rm :\longmapsto\:k = \dfrac{4 \: \pm \: 2 \sqrt{5} }{2}$

$\bf\implies \:k \: = \: 2 \: \pm \: \sqrt{5}$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$