Question

If the equation ${x^4} + 4{x^3} - 2{x^2} - 12x + 9 = 0$ has two pairs of equal roots, find the roots of the equation.

Answer 1

Hint: Find sum of roots and product of roots from the equation and then proceed.

Complete step-by-step answer:
From the question, the given equation is ${x^4} + 4{x^3} - 2{x^2} - 12x + 9 = 0$, and it has two pairs of equal roots.
Let the roots of the equation be $a,a,b$ and $b$. Then sum of roots is $2(a + b)$and product of roots is ${a^2}{b^2}.$
And we know, for any ${n^{th}}$degree polynomial:

${\text{Sum of roots }} = - \dfrac{{{\text{coefficient of }}{x^{n - 1}}}}{{{\text{coefficient of }}{x^n}}}$ and

${\text{Product of roots }} = \dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{x^n}}}$

In this case, the polynomial is of ${4^{th}}$degree. So,

${\text{Sum of roots }} = - \dfrac{{{\text{coefficient of }}{x^3}}}{{{\text{coefficient of }}{x^4}}} = - 4$ and

${\text{Product of roots }} = \dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{x^4}}} = 9$

Using these two results for the above equation, we'll get:
$
   \Rightarrow 2(a + b) = - 4, \\
   \Rightarrow a + b = - 2. \\
$
$
   \Rightarrow {a^2}{b^2} = 9 \\
   \Rightarrow ab = \pm 3. \\
$
Case $1$: $ab = 3,$
We know that, ${(a - b)^2} = {(a + b)^2} - 4ab.$ Using this, we’ll get:
$ \Rightarrow {(a - b)^2} = {( - 2)^2} - 4(3),$
$ \Rightarrow {(a - b)^2} = - 8$ which is not possible. Thus, $ab \ne 3.$
Case $2$: $ab = - 3,$
$
   \Rightarrow {(a - b)^2} = {(a + b)^2} - 4ab. \\
   \Rightarrow {(a - b)^2} = {( - 2)^2} - 4( - 3), \\
   \Rightarrow {(a - b)^2} = 16, \\
   \Rightarrow a - b = \pm 4 \\
$
So, we have $a + b = - 2,a - b = \pm 4.$ We will have two cases.
On adding these two equations, we’ll get:
$ \Rightarrow 2a = 2$ or $2a = - 6$
$ \Rightarrow a = 1$ or $a = - 3.$
Putting the value of $a$ in $a + b = - 2,$ we’ll get:
$b = - 3$ or $b = 1$
Thus the roots of the equation are $ - 3, - 3,1$ and $1.$

Note: For any ${n^{th}}$degree polynomial, $a{x^n} + b{x^{n - 1}} + c{x^{n - 2}} + d{x^{n - 3}} + ....... = 0$:
Sum of roots $ = - \dfrac{b}{a}$,
Sum of roots taking two at a time $ = \dfrac{c}{a}$,
Sum of roots taking three at a time $ = - \dfrac{d}{a}$, and so on.