Solving Quadratic Equations by Square Root Method: Basic Forms

TL;DR

You can solve basic quadratic equations that lack an "x" term by isolating the squared term and taking the square root of both sides. Remember to include both positive and negative roots. This method is quick when your equation looks like $ax^2 + c = 0$.

1. The Mental Model

Think of this method as "undoing" the squaring. Just like you'd subtract to undo addition, you take the square root to undo a square. The key is to get the $x^2$ part all by itself first.

2. The Core Material

The square root method is perfect for quadratic equations in the form $ax^2 + c = 0$ or $(x+k)^2 = d$. It's not suitable for equations that have an "x" term (like $bx$) because you can't easily isolate the squared part.

Isolating the Squared Term

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Your first step is always to get the term with $x^2$ (or $(x+k)^2$) by itself on one side of the equation. This often involves basic addition, subtraction, multiplication, or division.

For example, if you have $3x^2 - 12 = 0$:
1. Add 12 to both sides: $3x^2 = 12$
2. Divide by 3: $x^2 = 4$

Taking the Square Root

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Once you have $x^2 = \text{number}$ (or $(x+k)^2 = \text{number}$), you take the square root of both sides. This is the crucial step where you must remember both the positive and negative roots.

Why? Because both $2^2 = 4$ and $(-2)^2 = 4$. So, if $x^2 = 4$, then $x$ could be $2$ or $-2$. We write this as $x = \pm 2$.

If you're dealing with $(x+k)^2 = d$, then taking the square root gives you $x+k = \pm\sqrt{d}$. Then you'd just subtract $k$ from both sides to find $x$.

The Process

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Here's a simple flowchart of the steps:

graph TD
    A["Start with ax^2 + c = 0 or (x+k)^2 = d"] --> B["Isolate the squared term (x^2 or (x+k)^2)"]
    B --> C["Take the square root of BOTH sides"]
    C --> D["REMEMBER: Include positive ( + ) and negative ( - ) roots"]
    D --> E["Solve for x (if needed, e.g., if you have x+k)"]
    E --> F["You've found your two solutions for x!"]

Examples:

1. Simple $x^2 = \text{number}$ form:
   $x^2 = 25$
   $\sqrt{x^2} = \pm\sqrt{25}$
   $x = \pm 5$
   So, $x = 5$ or $x = -5$.

2. Needs isolating first:
   $2x^2 - 18 = 0$
   $2x^2 = 18$ (Add 18 to both sides)
   $x^2 = 9$ (Divide by 2)
   $\sqrt{x^2} = \pm\sqrt{9}$
   $x = \pm 3$
   So, $x = 3$ or $x = -3$.

3. With a quantity squared:
   $(x-3)^2 = 16$
   $\sqrt{(x-3)^2} = \pm\sqrt{16}$
   $x-3 = \pm 4$
   Now, you split this into two separate equations:
   a) $x-3 = 4 \implies x = 4+3 \implies x = 7$
   b) $x-3 = -4 \implies x = -4+3 \implies x = -1$
   So, $x = 7$ or $x = -1$.

3. Worked Example

Let's solve the equation $4(x+2)^2 - 100 = 0$.

1. Isolate the squared term:
   Start by adding 100 to both sides:
   $4(x+2)^2 = 100$

   Next, divide both sides by 4:
   $(x+2)^2 = \frac{100}{4}$
   $(x+2)^2 = 25$

2. Take the square root of both sides (remembering $\pm$):
   $\sqrt{(x+2)^2} = \pm\sqrt{25}$
   $x+2 = \pm 5$

3. Solve for $x$ by separating into two equations:

   • Case 1 (positive root):
     $x+2 = 5$
     $x = 5 - 2$
     $x = 3$

   • Case 2 (negative root):
     $x+2 = -5$
     $x = -5 - 2$
     $x = -7$

So, the solutions are $x = 3$ and $x = -7$.

4. Key Takeaways

• Use the square root method when the quadratic equation has an $x^2$ term and a constant term, but no simple $x$ term.
• Always isolate the $x^2$ term (or the quantity being squared) before taking the square root.
• When you take the square root of both sides, you must include both the positive and negative roots.
• If you end up with $x+k = \pm\text{number}$, remember to set up two separate equations to solve for $x$.
• This method is generally faster than factoring or the quadratic formula for these specific types of equations.

Common Mistakes to Avoid:
- Forgetting the "$\pm$" (plus or minus) when taking the square root, which means you'll only find one solution instead of two.
- Incorrectly isolating the squared term before taking the square root.
- Trying to use this method when there's an $x$ term (e.g., $x^2 + 5x + 6 = 0$) – it won't work easily.
- Making arithmetic errors when simplifying after taking the square root.

5. Now Try It

Solve the equation $3(x-1)^2 - 27 = 0$.

What to do: Follow the steps: isolate the squared term, take the square root (remembering $\pm$), and then solve for $x$.

What success looks like: You should find two distinct values for $x$.