Question

If one root of the equation 2x^2-x-2=0 is alpha, prove that the other root is 4(alpha)^3 - 6alpha - 3/2

Answer 1

Question

α is a solution of the equation, hence $2\alpha ^{2}-\alpha -2=0$.

We are given to prove $\beta =4\alpha ^{3}-6\alpha -\dfrac{3}{2}$.

Hint

Try eliminating the leading term.

Vieta's formula, $\alpha +\beta =-\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ for quadratic equation $ax^{2}+bx+c=0$.

Solution

In the first equation, we can write as $\alpha ^{2}=\dfrac{1}{2} \alpha +1$.

We try to eliminate the leading term

$4\alpha (\dfrac{1}{2} \alpha +1)-6\alpha -\dfrac{3}{2}$

$=2\alpha ^2+4\alpha -6\alpha -\dfrac{3}{2}$

$=2(\dfrac{1}{2} \alpha +1)-2\alpha -\dfrac{3}{2}$

$=\alpha +2-2\alpha -\dfrac{3}{2}$

$=-\alpha +\dfrac{1}{2}$

Now it is enough to prove that $\beta =-\alpha +\dfrac{1}{2}$. Since it is a quadratic equation, the sum of the roots is $\alpha +\beta =\dfrac{1}{2}$. This is equivalent to the given equation. $\blacksquare$

Answer 2

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given\::- }}}}$

The quadratic equation is 2x² - x - 2 = 0

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To \: prove \::- }}}}$

β = 4α³ - 6α - 3/2

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ How \: to \: Solve \::- }}}}$

Let the quadratic equation ax² + bx + c = 0.

• $\sf \alpha + \beta = - \dfrac{b}{a}$

• $\sf \alpha\beta = \dfrac{c}{a}$

• Write equation as 2α² - α - 2 (as α is a solution)

• Now , $\alpha^2 = \dfrac{1}{2}\alpha + 1$

$\dashrightarrow\qquad\sf Put \: \alpha^2 = \dfrac{1}{2}\alpha + 1 \:in \: \beta = 4\alpha^3 - 6\alpha - \dfrac{3}{2}$

$\dashrightarrow\qquad\sf= 4\alpha\bigg(\dfrac{1}{2}\alpha+1 \bigg)- 6\alpha - \dfrac{3}{2}$

$\dashrightarrow\qquad\sf = 2 \bigg(\dfrac{1}{2}\alpha + 1\bigg) - 2\alpha - \dfrac{3}{2}$

$\dashrightarrow\qquad\sf = \alpha + 2 - 2\alpha - \dfrac{3}{2}$

$\dashrightarrow\qquad\sf = - \alpha +\dfrac{1}{2}$

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Therefore,\::- }}}}$

• $\sf \beta = - \alpha + \dfrac{1}{2}$

• $\sf \alpha + \beta = \dfrac{1}{2}$

This is identical to the given equation.