Question

How do you solve the equation: $\left( {x - 3} \right)\left( {x + 7} \right) = 0$?

Answer 1

Hint: The given problem requires us to solve an equation. There are various methods that can be employed to solve a quadratic equation such as completing the square method, using quadratic formulas and by splitting the middle term. The given equation has already been given in the factored form. So, we don’t need to use any of the above mentioned methods as they are used only to factorize the equation and help in finding the roots of the equation.

Complete step by step solution:
In the given question, we are required to solve the given equation $\left( {x - 3} \right)\left( {x + 7} \right) = 0$.
Consider the equation $\left( {x - 3} \right)\left( {x + 7} \right) = 0$.
The equation could have been solved by various methods such as completing the square method, splitting the middle term and using the quadratic formula if it was given in the standard factored form of quadratic equation. Solving the quadratic equation by splitting the middle term, we have to split the middle term into two terms such that the sum of the terms gives us the original middle term and product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
However, the given equation is already in the factored form. Hence, we need to just equate it to zero and find the required roots of the equation.
$\left( {x - 3} \right)\left( {x + 7} \right) = 0$
Now, either $\left( {x - 3} \right) = 0$ or $\left( {x + 7} \right) = 0$.
So, either $x = 3$ and $x = \left( { - 7} \right)$.
Hence, the roots of the equation $\left( {x - 3} \right)\left( {x + 7} \right) = 0$ are \[x = \left( { - 7} \right)\] and $x = 3$.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as $2$. Quadratic equations can be solved by splitting the middle term, using the quadratic formula and completing the square method. Some equations don’t appear to be quadratic but can be reduced into quadratic equations using substitution.