Solving Quadratic Equations by Square Roots
Solve quadratic equations of the form ax² + c = 0 by isolating x² and taking the square root of both sides. Recognize that square root solutions come in ± pairs and simplify radical answers.
A lesson of Algebra 1, in Quadratic Functions, about 30 minutes.
Solving Quadratics by Square Roots
By the end of this lesson, you will be able to isolate $x^2$ and take a square root of both sides, remember the $\pm$ and know why it is there, simplify a radical answer, and recognize when there is no real solution.
Isolate the Square, Then Undo It
When a quadratic has no $x$ term, you do not need factoring at all.
$$x^2 - 25 = 0$$
$$x^2 = 25$$
$$x = \pm 5$$
Get $x^2$ alone, then square-root both sides. That is the whole method for this shape, and it is faster than anything else available.
Use it when $b = 0$, which covers $x^2 = k$, $ax^2 + c = 0$, and anything that can be rearranged into either. It also covers $(x - 3)^2 = 16$, where the squared thing is a bracket rather than a bare $x$, and that case turns out to matter enormously two lessons from now.
Why There Are Always Two
$$x^2 = 25 \quad \Rightarrow \quad x = 5 \text{ or } x = -5$$
Both work, because $(-5)^2 = 25$ just as $5^2 = 25$. Squaring destroys the sign, so undoing it has to restore both possibilities.
The $\sqrt{\ }$ symbol means the positive root only. $\sqrt{25}$ is $5$, not $\pm 5$, and that is a convention that keeps $\sqrt{\ }$ a function. So when you take a root of both sides of an EQUATION you must write the $\pm$ yourself:
$$x^2 = 25 \quad \Rightarrow \quad x = \pm\sqrt{25} = \pm 5$$
$$\sqrt{25} = 5 \quad \text{just an evaluation, no } \pm$$
Losing the $\pm$ throws away one of the two solutions, and on a graph it is the other place the parabola crosses the axis.
Example 1: The Whole Method
Here is every step on one equation, worked all the way through. $$3x^2 - 48 = 0$$
- Add 48 to both sides. What does the equation say now? 3x^2 = 48
- Divide both sides by 3. What does x squared equal? 16
- Take the square root of 16 4
- How many solutions does the equation have? 2
Simplifying a Radical Answer
Not every square root is whole, and the exact form is the correct answer.
$$x^2 = 12 \quad \Rightarrow \quad x = \pm\sqrt{12} = \pm 2\sqrt{3}$$
Pull out perfect squares. $12 = 4 \times 3$, and $\sqrt{4} = 2$, so $\sqrt{12} = 2\sqrt{3}$. The useful squares to look for are $4, 9, 16, 25, 36, 49, 64, 81, 100$.
Some to have at hand:
| Radical | Simplified | Because |
|---|---|---|
| $\sqrt{8}$ | $2\sqrt{2}$ | $8 = 4 \times 2$ |