Question

How do you solve $ \left( x+3 \right)\left( x-6 \right)=0 $ ?

Answer 1

Hint: In this problem we have to solve the given equation i.e., $ \left( x+3 \right)\left( x-6 \right)=0 $ . The solution of the equation is nothing but finding the values of $ x $ where the obtained $ x $ satisfies the given equation. For finding the solution for different kinds of equations we will follow different methods. In the given problem we have a quadratic equation that is factorized. Generally, to solve a quadratic equation we will either factorize it or use the direct formula that we have which is $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . But in this problem, we have the factorized quadratic equation, so there is no need of factorising it again. Now we will equate each term individually to zero and solve the obtained two equations to get the solution of the given equation.

Complete step by step answer:
Given that, $ \left( x+3 \right)\left( x-6 \right)=0 $ .
Equating each term individually to zero, then we will get
 $ x+3=0 $ or $ x-6=0 $ .
Considering the equation $ x+3=0 $ .
Subtracting $ 3 $ from both sides of the equation, then we will get
 $ x+3-3=-3 $
We know that $ +a-a=0 $ , then we will have
 $ x=-3....\left( \text{i} \right) $
Now considering the equation $ x-6=0 $ .
Adding $ 6 $ on both sides of the above equation, then we will get
 $ x-6+6=+6 $
We know that $ +a-a=0 $ , then we will have
 $ x=6....\left( \text{ii} \right) $
From equations $ \left( \text{i} \right) $ and $ \left( \text{ii} \right) $ we can write the solutions for the given equation as $ x=6,-3 $ .

Note:
In this problem, they have directly given the factorized form of a quadratic equation. So, we have directly equated each term to zero. Sometimes they only give the quadratic equation and ask to solve it. Then we need to factorize the given quadratic equation if possible. Otherwise, we will use the formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ to get the roots.