Question

How do you solve $ 40{x^3} - 10{x^2} - 5x = 0? $

Answer 1

Hint: First of all take the given expression and take the common multiple common from all the terms in the given equation and then find the roots of the equation or the value for “x” by splitting the middle term.

Complete step by step solution:
Take the given expression: $ 40{x^3} - 10{x^2} - 5x = 0 $
Take out the common multiple from all the terms in the above expression.
 $ 5x(8{x^2} - 2x - 1) = 0 $
Split the middle term in such a way that its product is equal to the product of the first and the last term.
\[
   - 2 = - 4 + 2 \\
   - 8 = 2( - 4) \;
 \]
Now, frame the equation accordingly
 $ 5x(8{x^2} - 4x + 2x - 1) = 0 $
Now make the pair of the first two and the last two terms.
 $ 5x(\underline {8{x^2} - 4x} + \underline {2x - 1} ) = 0 $
Find the common factors from the paired terms.
 $ \Rightarrow 5x[4x(2x - 1) + 1(2x - 1)] = 0 $
Take out the common factor common
 $ \Rightarrow 5x[(2x - 1)(4x + 1)] = 0 $
The above equation gives-
 $ \Rightarrow 5x = 0,\;{\text{2x - 1 = 0, 4x + 1 = 0}} $
Make the subject “x”. term multiplicative on one side if moved to the opposite side then it goes to the denominator. When any term in addition/subtraction is moved to the opposite side then its changes.
 $ \Rightarrow x = \dfrac{0}{5} $ or $ 2x = 1 $ or $ 4x = - 1 $
Simplify the equations, zero upon anything is always zero.
 $ \Rightarrow x = 0 $ or $ x = \dfrac{1}{2} $ or $ x = - \dfrac{1}{4} $
This is the required solution.
So, the correct answer is “ $ x = 0 $ or $ x = \dfrac{1}{2} $ or $ x = - \dfrac{1}{4} $ ”.

Note: Be careful about the sign convention while splitting the middle term. Here we were able to find the roots by splitting the middle term but in case roots are not real we won’t be able to do so, for that remember imaginary roots formula which can be expressed as \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\] and considering the general form of the quadratic equation $ a{x^2} + bx + c = 0 $ . Be careful about the sign convention and simplification of the terms in the equation.