Question

Find the quadratic equation whose roots are square of the roots of the equation x²+x+1​

Answer 1

Answer:

The quadratic equation is x2 + x + 1 = 0.

Answer 2

${x}^{2} + x + 1$

$\blue { \boxed{\bf roots =  \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} }}$

$\implies x = \frac{ - 1± \sqrt{ {1}^{2} - 4(1)(1)} }{2(1)}$

$\implies x = \frac{ - 1± \sqrt{ - 3}}{2}$

$\implies x = \frac{ - 1± \sqrt{ 3}i}{2}$

Since required equation's roots are square of roots of given equation.

$\implies x ^{2} = ( \frac{ - 1± \sqrt{ 3}i}{2} ) ^{2}$

$\implies x ^{2} = \frac{ 1 - 3± 2\sqrt{ 3}i}{4}$

$\implies x ^{2} = \frac{ - 2± 2\sqrt{ 3}i}{4}$

$\implies x ^{2} = \frac{ - 1± \sqrt{ 3}i}{2}$

$\therefore \alpha = \frac{ - 1 + \sqrt{3} i}{2}$

$\therefore \beta = \frac{ - 1 - \sqrt{3} i}{2}$

$\sf \purple{req. \: equation}$

${x}^{2} - ( \alpha + \beta ) x+ \alpha \beta$

$\implies {x}^{2} - ( \frac{ - 1 + \sqrt{3} i}{2} +\frac{ - 1 - \sqrt{3} i}{2} )x + ( \frac{ - 1 + \sqrt{3} i}{2} )( \frac{ - 1 - \sqrt{3} i}{2} )$

$\implies {x}^{2} - ( \frac{ - 1 + \sqrt{3} i - 1 - \sqrt{3} i}{2} )x + ( \frac{ { (- 1)}^{2} - ( \sqrt{3} i) ^{2} }{4} )$

$\implies {x}^{2} - ( \frac{ - 2}{2} )x + \frac{1- ( - 3) }{4}$

$\orange{ \therefore {x}^{2} + x + 1}$

HOPE IT HELPS!!