Question

How to solve $ 6{x^2} + x - 1 $ for $ x $ by factorizing?

Answer 1

Hint: The general form of a quadratic equation is \[a{x^2} + bx + c = 0\], where $ a \ne 0 $ . The values of $ x $ which satisfy the equation are called the roots or the solutions. These roots can be real or imaginary. In order to find the solution of a quadratic equation by way of factorization method, firstly we are required to write the given polynomial as the product of two linear factors which can be done by splitting the middle term and then by equating each factor to zero, we will get the desired roots or solutions of the given equation.

Complete step-by-step answer:
The general equation can be shown or written as \[{x^2} + Sx + P = 0\]where S = sum of roots and P = product of roots. The given equation is $ 6{x^2} + x - 1 = 0 $ . We need to find two factors of $ - 6 $ , such that the sum of those two factors is $ 1 $ and the product of those two factors is $ - 6 $ .
 $
   \Rightarrow 6{x^2} + x - 1 = 0 \\
   \Rightarrow 6{x^2} + 3x - 2x - 1 = 0 \;
  $
Now, we take out the common factors from the two pairs
 $
   \Rightarrow 3x\left( {2x + 1} \right) - 1\left( {2x + 1} \right) = 0 \\
   \Rightarrow \left( {3x - 1} \right)\left( {2x + 1} \right) = 0 \;
  $
Now, we have to equate $ \left( {3x - 1} \right) $ and $ \left( {2x + 1} \right) $ with zero in order to find the roots of the given quadratic equation.
 $
   \Rightarrow 3x - 1 = 0 \\
   \Rightarrow 3x = 1 \\
   \Rightarrow x = \dfrac{1}{3} \;
  $ $
   \Rightarrow 2x + 1 = 0 \\
   \Rightarrow 2x = - 1 \\
   \Rightarrow x = \dfrac{{ - 1}}{{\;2}} \;
  $
Hence, the roots of the given quadratic equation are $ \dfrac{1}{3} $ and $ \dfrac{{ - 1}}{2} $
So, the correct answer is “ $ \dfrac{1}{3} $ and $ \dfrac{{ - 1}}{2} $ ”.

Note: As we have now found the roots of the equation, there is a way to double-check our solution. In order to double-check the answer all, we have to substitute the value of $ x $ one by one in the given quadratic equation. If after substituting the value of $ x $ we get zero i.e., LHS=RHS then we know for sure that our solution is correct.