Question

If one root of the equation x^2-x-k=0 is square of the other, then find k!?? ​

Answer 1

Question:

If one root of the equation x² - x - k=0 is square of the other, then find k ?

To find:

★ The value of k.

Answer:

The value of $\sf\pink{ \bold{k \:is\: 2± \sqrt{ 5}}}$

Given:

★ An equation is given,

${x}^{2} - x - k = 0$

Step-by-step explanation:

Given that, One root is square of the other ;

Let one root be $\alpha$

Let the other root be ${\alpha}^{2}$

${x}^{2} - x - k = 0$

It's of the form, $a{x}^{2}+bx+c=0$

Where;

a = 1, b = -1, c = -k

We know that,

$\boxed{Sum \: of \: roots = \frac{ - b}{a} }$

$\implies \alpha + \beta = \frac{ - ( - 1)}{1}$

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

Now,

$\leadsto {( \alpha + { \alpha }^{2} )}^{3} = 1$

$\leadsto { \alpha }^{3} + { \alpha }^{6} + 3 { \alpha }^{3} ( \alpha + { \alpha }^{2}) --->(1)$

$\boxed{Product \: of \: roots = \frac{ c}{a} }$

$\longmapsto \alpha \beta = \frac{ - k}{1}$

$\longmapsto \alpha {( \alpha )}^{2} = \frac{ - k}{1}$

$\longmapsto { \alpha }^{3} = - k$

Now substituting the value of ${\alpha}^{3}$ in eq (1)

$\hookrightarrow - k + {k}^{2} + 3( - k)(1) = 1$

$\hookrightarrow {k}^{2} - 4 k - 1 =0$

Now factorization by formula method:-

$\boxed{k = \frac{ - b± \sqrt{ {b}^{2} - 4ac \: } }{2} }$

$\hookrightarrow k = \frac{4± \sqrt{ 16 + 4} }{2}$

$\hookrightarrow \boxed{k = 2± \sqrt{ 5} }$

$\therefore$The value of k is $\bold{2± \sqrt{ 5}}$

Formula used:

$\bigstar Sum \: of \: roots = \frac{ - b}{a}$

$\bigstar Product \: of \: roots = \frac{ c}{a}$

$\bigstar k = \dfrac{ - b± \sqrt{ {b}^{2} - 4ac \: } }{2}$

Hint given in the question :

★ One root is square of the other .

One root be $\alpha$

Other root be ${\alpha}^{2}$

⚠️Note⚠️

We must read the hint properly because it forms the equation.