Question

If one root of \[{x^2} - x - k = 0\] maybe the square of the other, then k =
A) $2 \pm \sqrt 5 $
B) $2 \pm \sqrt 3 $
C) $3 \pm \sqrt 2 $
D) $5 \pm \sqrt 2 $

Answer 1

Hint: First find the relation between roots and the coefficients. Now take the cube of the sum of the roots. Then expand the cube by the formula ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$. Then substitute the values from the sum of the roots and product of the roots which will make a quadratic equation. After that solve the equation by the quadratic formula $k = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. The value came after simplification is the desired result.

Complete step by step answer:
Given: - One root of the equation is the square of the other root.
Let the roots of the equation be $\alpha ,{\alpha ^2}$.
The relation between the roots and the coefficients are,
$\alpha + \beta = - \dfrac{b}{a}$
\[\alpha \beta = \dfrac{c}{a}\]
Substitute the values,
$ \Rightarrow \alpha + {\alpha ^2} = - \dfrac{{\left( { - 1} \right)}}{1}$
\[ \Rightarrow \alpha \left( {{\alpha ^2}} \right) = \dfrac{{ - k}}{1}\]
Simplify the terms,
$ \Rightarrow \alpha + {\alpha ^2} = 1$ …… (1)
$ \Rightarrow {\alpha ^3} = - k$ …… (2)
Now, take the cube of $\alpha + {\alpha ^2} = 1$,
$ \Rightarrow {\left( {\alpha + {\alpha ^2}} \right)^3} = {1^3}$
As we know,
${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
Substitute the values,
$ \Rightarrow {\alpha ^3} + {\left( {{\alpha ^2}} \right)^3} + 3\alpha \times {\alpha ^2}\left( {\alpha + {\alpha ^2}} \right) = 1$
Simplify the terms,
$ \Rightarrow {\alpha ^3} + {\alpha ^6} + 3{\alpha ^3}\left( {\alpha + {\alpha ^2}} \right) = 1$
Substitute the values from equation (1) and (2),
$ \Rightarrow - k + {\left( { - k} \right)^2} + 3 \times - k \times 1 = 1$
Multiply the terms,
$ \Rightarrow {k^2} - 4k = 1$
Move 1 to the left side,
$ \Rightarrow {k^2} - 4k - 1 = 0$
Now, solve by quadratic formula,
$k = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substitute the values,
$ \Rightarrow k = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times 1 \times - 1} }}{{2 \times 1}}$
Multiply the terms,
$ \Rightarrow k = \dfrac{{4 \pm \sqrt {16 + 4} }}{2}$
Add the terms in the square root,
$ \Rightarrow k = \dfrac{{4 \pm \sqrt {20} }}{2}$
Take common factors out from the square root,
$ \Rightarrow k = \dfrac{{4 \pm 2\sqrt 5 }}{2}$
Cancel out the common factors,
$ \Rightarrow k = 2 \pm \sqrt 5 $
Thus, the value of k is $2 \pm \sqrt 5 $.

Hence, option (A) is the correct answer.

Note:
A quadratic equation is a polynomial equation of degree 2. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution, or two imaginary solutions.
There are several methods you can use to solve a quadratic equation:
- Factoring
- Completing the Square
- Quadratic Formula
- Graphing
All methods start with setting the equation equal to zero.