Solving Quadratic Equations by the Square Root Rule

TL;DR

You can solve specific types of quadratic equations by isolating the squared term and then taking the square root of both sides. Remember to always consider both positive and negative roots. This method is quickest when your quadratic equation has no linear 'x' term.

1. The Mental Model

Think of this method as undoing the squaring operation. If you have something squared equals a number, you can find that "something" by taking the square root of the number. It's like working backward from a perfect square.

2. The Core Material

The square root rule is ideal for quadratic equations that look like $ax^2 + c = 0$ or $(x+k)^2 = d$. The key is that there's no single 'x' term (e.g., no $bx$ term).

How It Works

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1. Isolate the squared term: Get the $x^2$ or $(x \pm k)^2$ part by itself on one side of the equation.
2. Take the square root of both sides: Remember that the square root of a number has both a positive and a negative solution. So, $\sqrt{d}$ will give you $\pm\sqrt{d}$.
3. Solve for x: If you have $(x \pm k) = \pm\sqrt{d}$, you'll need one more step to get x alone.

Let's look at the process:

graph TD
    A["Start with ax² + c = 0 or (x+k)² = d"] --> B["Isolate the squared term"]
    B --> C{"Is the right side positive, negative, or zero?"}
    C -- "Negative" --> D["No real solutions (imaginary solutions)"]
    C -- "Zero" --> E["One solution: x = -k (if (x+k)²=0) or x=0 (if x²=0)"]
    C -- "Positive" --> F["Take square root of both sides (remember ±)"]
    F --> G["Solve for x (two distinct real solutions)"]
    G --> H["End"]
    E --> H
    D --> H

Example 1: Simple Case ($ax^2 + c = 0$)

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Let's solve $2x^2 - 18 = 0$.

1. Isolate the squared term:
   $2x^2 = 18$
   $x^2 = \frac{18}{2}$
   $x^2 = 9$

2. Take the square root of both sides:
   $\sqrt{x^2} = \pm\sqrt{9}$
   $x = \pm3$

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

Example 2: More Complex Case ($(x+k)^2 = d$)

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Let's solve $(x-5)^2 = 16$.

1. The squared term is already isolated! $(x-5)^2 = 16$

2. Take the square root of both sides:
   $\sqrt{(x-5)^2} = \pm\sqrt{16}$
   $x-5 = \pm4$

3. Solve for x:
   We now have two separate linear equations to solve:

   • $x-5 = 4$
     $x = 4+5$
     $x = 9$
   • $x-5 = -4$
     $x = -4+5$
     $x = 1$

So, the solutions are $x = 9$ and $x = 1$.

Important Note: No Real Solutions

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What if you end up with something like $x^2 = -9$?
If you take the square root of both sides, you get $x = \pm\sqrt{-9}$. In the realm of real numbers, you can't take the square root of a negative number. So, in this case, there are no real solutions. (You'll learn about imaginary numbers later, which would provide solutions here.)

3. Worked Example

Let's solve the equation $3(x+2)^2 - 27 = 0$.

1. Isolate the squared term $(x+2)^2$:
   $3(x+2)^2 = 27$
   $(x+2)^2 = \frac{27}{3}$
   $(x+2)^2 = 9$

2. Take the square root of both sides:
   $\sqrt{(x+2)^2} = \pm\sqrt{9}$
   $x+2 = \pm3$

3. Solve for x:

   • Case 1: $x+2 = 3$
     $x = 3 - 2$
     $x = 1$
   • Case 2: $x+2 = -3$
     $x = -3 - 2$
     $x = -5$

The solutions are $x=1$ and $x=-5$.

4. Key Takeaways

• This method is best for quadratic equations without an 'x' term ($ax^2 + c = 0$) or those already in the form $(x \pm k)^2 = d$.
• Always isolate the squared term first before taking the square root.
• Remember to include both the positive and negative square roots ($\pm$) to find all possible solutions.
• If you end up taking the square root of a negative number, there are no real solutions.
• The two solutions are often symmetric around a central point, or one of them.
• This method is typically faster than factoring or the quadratic formula when applicable.

5. Now Try It

Solve the equation $4(x-1)^2 - 100 = 0$. What are the two values for 'x' that satisfy this equation? Success looks like finding both correct real solutions.