Question

How to solve this? Find $ m \in R $ for which $ {x^3} - 3x + m = 0 $ has a double root.

Answer 1

Hint: Here the given equation is in the form of the cubic equation. So, there are three roots of the equation also given that it has a double root so we will assume two same roots and a third root different and solve accordingly to find the value for “m”.

Complete step by step solution:
Take the given equation: $ {x^3} - 3x + m = 0 $
As we know that cubic equation has three roots but given that it contains double root, so let us assume that the roots of the above equation are $ \alpha ,\alpha $ and $ \beta $
Now, we Can write the given equation as-
 $ {x^3} - 3x + m = (x - \alpha )(x - \alpha )(x - \beta ) $
Expand the equation on the right hand side of the equation one by one, first multiply first two terms.
 $ {x^3} - 3x + m = {(x - \alpha )^2}(x - \beta ) $
Apply the identity: $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ in the above equation on the right hand side of the equation.
 $ {x^3} - 3x + m = ({x^2} - 2x\alpha + {\alpha ^2})(x - \beta ) $
Multiply both the brackets on the right hand side of the equation. Remember when there is a negative sign outside the bracket then the sign of the term inside the bracket also changes. Positive terms become negative and vice-versa.
\[
  {x^3} - 3x + m = ({x^2}(x - \beta ) - 2x\alpha (x - \beta ) + {\alpha ^2}(x - \beta )) \\
  {x^3} - 3x + m = ({x^3} - {x^2}\beta - 2{x^2}\alpha + 2x\alpha \beta + {\alpha ^2}x - {\alpha ^2}\beta ) \;
 \]
Bring like terms together and simplify-
\[{x^3} - 3x + m = {x^3} - (2\alpha + \beta ){x^2} + ({\alpha ^2} + 2\alpha \beta )x - {\alpha ^2}\beta \]
Compare, both the sides of the equation-
 $
   \Rightarrow 2\alpha + \beta = 0\;{\text{ }}....{\text{ (I)}} \\
  {\alpha ^2} + 2\alpha \beta = - 3\;{\text{ }}....{\text{ (II)}} \\
   - {\alpha ^2}\beta = m\;{\text{ }}....{\text{ (III)}} \;
  $
From the equation (I)
 $ \Rightarrow \beta = - 2\alpha $ …. (IV)
Place the above value in the equation (II)
 $ \Rightarrow - 3{\alpha ^2} = - 3 $
Common factors from both the sides of the cancel each other.
 $ \Rightarrow {\alpha ^2} = 1 $
Take square root on both the sides of the equation-
 $ \Rightarrow \alpha = \pm 1 $ …. (V)
And placing value of equation (IV) in Equation (III) gives
 $ 2{\alpha ^3} = m $
 $ \Rightarrow m = 2( \pm 1) $
Simplification implies –
 $ \Rightarrow m = \pm 2 $
This is the required solution.
So, the correct answer is “ $ m = \pm 2 $ ”.

Note: In algebra, a cubic equation in one variable is an equation of the form.
The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. ...
The coefficients do not need to be real numbers.
Be careful about the sign convention when you open the brackets. When there is a negative sign outside the bracket then the sign of the terms inside the bracket changes when the bracket is opened while when there is a positive term outside the bracket then there is no change in the sign of the terms inside the bracket.