Solving Quadratic Equations by Extracting Square Roots
July 12, 2025

Introduction
A quadratic equation is a polynomial equation of the form:
where a, \, b, \, \text{ and } c are real numbers and a\neq0.
- The term ax^2 is called the quadratic term.
- The term bx is the linear term.
- The term c is the constant term.
We will be focusing on finding the solutions to quadratic equations, starting by extracting square roots. Before we proceed, let us understand first the Square Root Property, which will be our guideline.
Quadratic equations that can be written in the form x^2 = k can be solved using the following rules:
- If k > 0, \text{ then } x^2 = k has two real solutions or roots: x = \pm \sqrt{k}.
- If k=0 then x^2 = k has one real solution or root: x = 0.
- If k < 0 then x^2 = k has no real solutions or roots.
Illustrative Example #1
Solve \bm{x^2 = 36}
Since 36 > 0, then it has two solutions or roots. Thus,
Illustrative Example #2
Solve \bm{m^2 - 49 = 0}
Since 49 > 0, then it has two solutions or roots. Thus,
Illustrative Example #3
Solve \bm{y^2 - 5 = 0}
Since 5 > 0, then it has two solutions or roots. Thus,
Illustrative Example #4
Solve \bm{t^2 = 50}
Since 50 > 0, then it has two solutions or roots. Thus,
Illustrative Example #5
Solve \bm{r^2 + 100 = 0}
First, isolate the squared variable:
\begin{aligned} r^2 + 100 &= 0 \\ r^2 &= -100 \end{aligned}
Since the square of a real number cannot be negative, the equation has no real solutions. Thus, \boxed{\text{ No real roots}}.
Illustrative Example #6
Solve \bm{(x - 9)^2 = 64}
Since 64 > 0, then it has two solutions or roots. Thus,
Therefore, the solutions are
Illustrative Example #7
Solve \bm{(x - 4)^2 = 36}
Since 36 > 0, then it has two solutions or roots. Thus,
Therefore, the solutions are
Illustrative Example #8
Solve \bm{(2x + 3)^2 = 50}
Since 50 > 0, then it has two solutions or roots. Thus,
Therefore, the solutions are
Illustrative Example #9
Solve \displaystyle \bm{\left(x + \frac{1}{2}\right)^2 = 0}
Since the right-hand side is 0, there is only one real solution. Thus,
Therefore, the solution is \boxed{x = -\dfrac{1}{2}}
Illustrative Example #10
Solve \bm{(x - 3)^2 + 18 = 0}
First, isolate the squared term:
Since the square of a real number cannot be negative, the equation has no real solutions. Thus, \boxed{\text{No real roots}}.