Pure Quadratic Equations
Understanding pure quadratic equations
A pure quadratic equation is a second-degree equation of the form ax2+bx+c=0, where the middle term (b) is zero, while a and c are both non-zero. $$ ax^2 + c = 0 $$
These equations are straightforward to solve. You don't need the general quadratic formula. Instead, you can isolate x directly and express it in terms of a and c.
$$ x^2 = \frac{-c}{a} $$
Since x is squared, the next step is to take the square root of both sides.
$$ \sqrt{x^2} = \sqrt{ \frac{-c}{a} } $$
$$ x = \pm \sqrt{ \frac{-c}{a} } $$
This leads to two possible situations:
- When a and c have the same sign
If both coefficients are positive or both are negative, the equation has no real solutions. In this case, the expression under the square root (-c/a) is negative, and no real number squared gives a negative result. - When a and c have opposite signs
If one coefficient is positive and the other negative, the equation has two distinct real solutions, because the value under the square root (-c/a) is positive.
Note. In a pure quadratic equation, both a and c must be non-zero. If either were zero, it would no longer fit the definition of a pure quadratic equation.
Example 1
Let's solve this pure quadratic equation:
$$ 2x^2 - 8 = 0 $$
The coefficients are a=2 and c=-8.
Since a and c have opposite signs, the equation has two distinct real solutions.
First, isolate x by moving constants to the other side:
$$ 2x^2 - 8 + 8 = 0 + 8 $$
$$ 2x^2 = 8 $$
Divide both sides by 2:
$$ \frac{2x^2}{2} = \frac{8}{2} $$
$$ x^2 = 4 $$
Now take the square root of both sides:
$$ \sqrt{x^2} = \sqrt{4} $$
The square root cancels the exponent on the left, giving:
$$ x = \pm 2 $$
So the equation has two real solutions: x1=-2 and x2=2.
$$ x = \begin{cases} x_1 = -2 \\ \\ x_2 = 2 \end{cases} $$
Example 2
Now consider another pure quadratic equation:
$$ 4x^2 + 8 = 0 $$
Here, a=4 and c=8.
Because both coefficients have the same sign, the equation has no real solutions.
Verification. Isolate x by moving the constant term to the other side. $$ 4x^2 = -8 $$ Divide both sides by 4: $$ 4x^2 \cdot \frac{1}{4} = -8 \cdot \frac{1}{4} $$ $$ x^2 = -2 $$ Then take the square root of both sides: $$ \sqrt{x^2} = \sqrt{-2} $$ $$ x = \pm \sqrt{-2} $$ Since the number inside the square root is negative, there are no real solutions.
Once you understand this reasoning, you can solve any pure quadratic equation simply by examining the signs of the coefficients and applying the basic properties of square roots.
