9.1 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 the right form. Instead of factoring or using the quadratic formula, you isolate the squared term and take the square root of both sides.

This section covers how to apply the Square Root Property to two equation forms: $ax^2 = k$ and $a(x - h)^2 = k$. You'll also learn how to interpret the types of solutions you get.

Solving Quadratic Equations Using the Square Root Property

Square Root Property for ax² = k

The Square Root Property says: if $x^2 = d$, then $x = \pm\sqrt{d}$. The $\pm$ is there because both a positive and a negative number, when squared, give the same result. For instance, both $3$ and $-3$ square to $9$.

To solve an equation in the form $ax^2 = k$:

1. Divide both sides by $a$ to isolate $x^2$: $x^2 = \frac{k}{a}$
2. Take the square root of both sides, using $\pm$: $x = \pm\sqrt{\frac{k}{a}}$
3. Simplify the radical if possible (look for perfect square factors)

The expression under the square root sign is called the radicand. If the radicand is a perfect square, your answers will be integers. If not, leave the answer in simplest radical form or approximate as a decimal.

Example 1: Solve $4x^2 = 36$

$x^2 = 9 \rightarrow x = \pm\sqrt{9} \rightarrow x = \pm 3$

Example 2: Solve $5x^2 = 60$

$x^2 = 12 \rightarrow x = \pm\sqrt{12} = \pm 2\sqrt{3} \approx \pm 3.464$

Square Root Property with a(x - h)²

When the squared expression contains a variable term like $(x - h)$, the same property applies. You just have one extra step at the end: solve for $x$ after taking the square root.

1. Divide both sides by $a$ to isolate the squared binomial: $(x - h)^2 = \frac{k}{a}$

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

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

4. Write out both solutions separately

Example 1: Solve $3(x - 2)^2 = 75$

$(x - 2)^2 = 25 \rightarrow x - 2 = \pm 5 \rightarrow x = 2 + 5 = 7 \text{ or } x = 2 - 5 = -3$

Example 2: Solve $2(x + 1)^2 = 18$

Note that $(x + 1)$ is the same as $(x - (-1))$, so $h = -1$.

$(x + 1)^2 = 9 \rightarrow x + 1 = \pm 3 \rightarrow x = -1 + 3 = 2 \text{ or } x = -1 - 3 = -4$

A common mistake here is forgetting the $\pm$ and only writing one solution. Always check that you've found two values (unless the radicand is zero).

The value of $h$ also has a graphical meaning: it's the x-coordinate of the vertex (and the axis of symmetry) of the parabola.

Interpreting Quadratic Equation Solutions

The type of number you get under the square root tells you a lot about the solutions:

• Positive perfect square radicand → two rational solutions (nice integers or fractions). Example: $x^2 = 25$ gives $x = \pm 5$
• Positive non-perfect-square radicand → two irrational solutions (leave in radical form or approximate). Example: $x^2 = 7$ gives $x = \pm\sqrt{7} \approx \pm 2.646$
• Radicand equals zero → one solution (a double root). Example: $(x - 4)^2 = 0$ gives $x = 4$
• Negative radicand → no real solutions; instead you get two complex solutions involving $i$, where $i = \sqrt{-1}$. Example: $x^2 = -9$ gives $x = \pm 3i$

These solution types connect directly to the graph of the quadratic function:

• Two real solutions → the parabola crosses the x-axis at two points
• One real solution (double root) → the parabola touches the x-axis at its vertex
• No real solutions → the parabola doesn't reach the x-axis at all

If you've seen the discriminant ($b^2 - 4ac$) from the standard form $ax^2 + bx + c = 0$, it tells you the same information. A positive discriminant means two real roots, zero means one, and negative means complex roots. You'll use the discriminant more when you get to the quadratic formula, but the underlying idea is the same.

Additional Methods and Considerations

The Square Root Property works well when the equation is already in the form $ax^2 = k$ or $a(x - h)^2 = k$. If the equation has a linear $bx$ term that you can't easily handle by rewriting, you'll need other methods like factoring, completing the square, or the quadratic formula.

When simplifying radicals, always check whether the radicand has perfect square factors. For example, $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$.

Finally, remember that the graph of any quadratic equation is a parabola. The solutions you find with the Square Root Property correspond to the x-intercepts of that parabola, which is why they're also called roots or zeros of the function.