Solving Quadratics by Taking the Square Root: A Khan Academy Deep Dive

Welcome to another exploration of quadratic equations! In this post, we'll focus on a specific technique: solving quadratics by isolating the squared term and then taking the square root. This method is particularly effective when the quadratic equation is in a form that allows for easy isolation. We'll use examples inspired by the Khan Academy topic to illustrate the process.

When to Use This Method

The "taking the square root" method shines when your quadratic equation looks something like this:

$$ax^2 + c = 0$$

Notice that there's no $x$ term (no term with $x$ raised to the power of 1). This absence is key! When the equation lacks a linear term (the 'bx' term in the general quadratic form $ax^2 + bx + c = 0$), isolating the $x^2$ term becomes straightforward.

The Steps Involved

Let's break down the process into manageable steps:

  1. Isolate the Squared Term: Use algebraic manipulation to get the $x^2$ term by itself on one side of the equation. This usually involves adding or subtracting constants from both sides and then dividing by the coefficient of $x^2$.
  2. Take the Square Root of Both Sides: Once the $x^2$ term is isolated, take the square root of both sides of the equation. Remember the golden rule: Always consider both the positive and negative square roots! This is because both a positive number and its negative counterpart, when squared, result in the same positive number.
  3. Solve for x: Simplify the resulting expressions to find the possible values of $x$.

Khan Academy Examples: A Closer Look

Let's consider some examples similar to those found on Khan Academy:

Example 1:

Solve for x: $x^2 - 9 = 0$

  1. Isolate $x^2$: Add 9 to both sides: $x^2 = 9$
  2. Take the square root: $\sqrt{x^2} = \pm\sqrt{9}$
  3. Solve for x: $x = \pm 3$. So, $x = 3$ or $x = -3$

Example 2:

Solve for x: $2x^2 - 50 = 0$

  1. Isolate $x^2$: Add 50 to both sides: $2x^2 = 50$. Divide both sides by 2: $x^2 = 25$
  2. Take the square root: $\sqrt{x^2} = \pm\sqrt{25}$
  3. Solve for x: $x = \pm 5$. So, $x = 5$ or $x = -5$

Example 3:

Solve for x: $(x + 2)^2 = 16$

  1. The squared term is already isolated.
  2. Take the square root: $\sqrt{(x + 2)^2} = \pm\sqrt{16}$
  3. Simplify: $x + 2 = \pm 4$
  4. Solve for x: $x = -2 \pm 4$. So, $x = -2 + 4 = 2$ or $x = -2 - 4 = -6$

Important Considerations

  • The $\pm$ Sign: Never forget the $\pm$ sign when taking the square root. It represents two possible solutions.
  • No Real Solutions: If, after isolating the $x^2$ term, you end up with $x^2$ equal to a negative number, there are no real solutions. The square of any real number is always non-negative. The solutions would be imaginary numbers.

Practice Makes Perfect!

The best way to master this technique is through practice. Work through more examples on Khan Academy and other resources. Don't be afraid to make mistakes – they're a crucial part of the learning process. Good luck, and happy solving!