10.1 Solve Quadratic Equations Using the Square Root Property

Quadratic equations show up constantly in algebra, and the square root property gives you a direct way to solve them when the equation has a specific structure: a squared term equal to a constant, like $ax^2 = k$ or $(x - h)^2 = k$.

Instead of factoring or using the quadratic formula, this method works by isolating the squared expression and taking the square root of both sides. It's especially powerful when you can recognize perfect square trinomials and rewrite them as the square of a binomial.

Solving Quadratic Equations Using the Square Root Property

Square root property for $ax^2 = k$

The square root property says: if $x^2 = c$, then $x = \pm\sqrt{c}$. The $\pm$ is critical because both a positive and a negative number, when squared, give the same result. For example, both $3^2$ and $(-3)^2$ equal 9.

Here's the process:

1. Isolate the squared term on one side of the equation (divide or move constants as needed).
2. Take the square root of both sides.
3. Write the solution with $\pm$ in front of the simplified square root.

Example: Solve $4x^2 = 36$

1. Divide both sides by 4 to isolate $x^2$: $x^2 = 9$
2. Take the square root of both sides: $x = \pm\sqrt{9}$
3. Simplify: $x = \pm 3$

So $x = 3$ or $x = -3$. Graphically, these are the x-coordinates where the parabola $y = 4x^2 - 36$ crosses the x-axis.

Square root property for $(x - h)^2 = k$

When the squared part is a binomial like $(x - h)^2$, the same property applies. You just have one extra step at the end: solve for $x$ by isolating it.

1. Take the square root of both sides: $x - h = \pm\sqrt{k}$

2. Simplify the square root if possible.

3. Add $h$ to both sides to solve for $x$: $x = h \pm\sqrt{k}$

Example: Solve $(x - 2)^2 = 16$

1. Take the square root of both sides: $x - 2 = \pm\sqrt{16}$

2. Simplify: $x - 2 = \pm 4$

3. Add 2 to both sides: $x = 2 \pm 4$

This gives two solutions: $x = 2 + 4 = 6$ or $x = 2 - 4 = -2$.

Perfect square trinomials and the square root property

A perfect square trinomial is a trinomial that factors neatly into the square of a binomial. Recognizing these lets you convert a trinomial equation into the $(x - h)^2 = k$ form and then apply the square root property.

The two patterns are:

• $a^2 + 2ab + b^2 = (a + b)^2$
• $a^2 - 2ab + b^2 = (a - b)^2$

To check whether a trinomial is a perfect square:

• Are the first and last terms perfect squares?
• Is the middle term equal to twice the product of the square roots of the first and last terms?

If both answers are yes, you can factor it.

Example: Solve $x^2 + 6x + 9 = 25$

1. Recognize the left side as a perfect square trinomial: $x^2 + 6x + 9 = (x + 3)^2$

   • $x^2$ is a perfect square, $9 = 3^2$ is a perfect square, and $6x = 2 \cdot x \cdot 3$ ✓

2. Rewrite the equation: $(x + 3)^2 = 25$

3. Take the square root of both sides: $x + 3 = \pm 5$

4. Solve for $x$: $x = -3 \pm 5$

This gives $x = -3 + 5 = 2$ or $x = -3 - 5 = -8$.

Additional Solving Methods for Quadratic Equations

The square root property is one of several ways to solve quadratic equations (equations where the highest exponent is 2):

• Factoring works when the quadratic expression can be written as a product of two linear factors, like $(x + 2)(x - 5) = 0$.
• The quadratic formula ($x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$) works for any quadratic equation, even when factoring isn't easy and the square root property doesn't directly apply.

The square root property is typically the fastest method when the equation is already in the form $ax^2 = k$ or $(x - h)^2 = k$, so learn to spot those structures quickly.