Solving Quadratics by Square Roots

Not every quadratic equation needs factoring. When the equation has a squared term but no linear term (no plain $x$ by itself), you can solve it directly by isolating the squared expression and taking the square root of both sides. This method is faster than factoring and does not require the quadratic formula — it is the most direct route to the answer for the right type of problem.

The square root method is also essential preparation for completing the square, which you will learn next.

When to Use the Square Root Method

Use this method when the equation fits one of these patterns:

• $x^2 = k$ (pure quadratic — no $x$ term)
• $ax^2 = k$ (coefficient on $x^2$ but still no $x$ term)
• $(x - h)^2 = k$ (squared binomial equals a constant)
• $a(x - h)^2 = k$ (coefficient on the squared binomial)

The common feature: the variable appears only inside a squared expression, with no separate linear $x$ term.

The Key Principle: The $\pm$ Symbol

When you take the square root of both sides of an equation, you must account for both the positive and negative roots:

$\text{If } x^2 = k, \text{ then } x = \sqrt{k} \text{ or } x = -\sqrt{k}$

We write this compactly as $x = \pm\sqrt{k}$.

Why both? Because $5^2 = 25$ and $(-5)^2 = 25$. Squaring erases the sign, so when you reverse the process, you must restore both possibilities.

Forgetting the $\pm$ is the single most common mistake with this method. It will cost you one of your two solutions every time.

Solving $x^2 = k$

Example 1: $x^2 = 36$

Take the square root of both sides:

$x = \pm\sqrt{36}$

$x = \pm 6$

Answer: $x = 6$ or $x = -6$

Check: $6^2 = 36$ . $(-6)^2 = 36$ . Both work.

Example 2: $x^2 = 20$

Take the square root of both sides:

$x = \pm\sqrt{20}$

Simplify the radical: $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$.

$x = \pm 2\sqrt{5}$

As decimals, $x \approx \pm 4.47$.

Answer: $x = 2\sqrt{5}$ or $x = -2\sqrt{5}$

Example 3: $x^2 = -9$

There is no real number whose square is negative. The square of any real number is zero or positive.

Answer: No real solutions. (You will encounter imaginary numbers in Algebra 2, where the answer would be $x = \pm 3i$, but for Algebra 1, this equation simply has no solution.)

Solving $ax^2 = k$

When there is a coefficient on $x^2$, isolate $x^2$ first by dividing both sides by that coefficient.

Example 4: $3x^2 = 75$

Step 1 — Divide both sides by 3:

$x^2 = 25$

Step 2 — Take the square root of both sides:

$x = \pm 5$

Answer: $x = 5$ or $x = -5$

Example 5: $5x^2 - 80 = 0$

Step 1 — Add 80 to both sides:

$5x^2 = 80$

Step 2 — Divide both sides by 5:

$x^2 = 16$

Step 3 — Take the square root:

$x = \pm 4$

Answer: $x = 4$ or $x = -4$

Solving $(x - h)^2 = k$

This is where the square root method becomes especially powerful. You can handle squared binomials without expanding them.

Example 6: $(x - 3)^2 = 16$

Step 1 — Take the square root of both sides:

$x - 3 = \pm 4$

Step 2 — Solve both cases:

$x - 3 = 4 \implies x = 7$

$x - 3 = -4 \implies x = -1$

Check $x = 7$: $(7 - 3)^2 = 4^2 = 16$ . Correct.

Check $x = -1$: $(-1 - 3)^2 = (-4)^2 = 16$ . Correct.

Answer: $x = 7$ or $x = -1$

Example 7: $(x + 5)^2 = 12$

Step 1 — Take the square root of both sides:

$x + 5 = \pm\sqrt{12} = \pm 2\sqrt{3}$

Step 2 — Subtract 5 from both sides:

$x = -5 \pm 2\sqrt{3}$

As decimals: $x \approx -5 + 3.46 = -1.54$ or $x \approx -5 - 3.46 = -8.46$.

Answer: $x = -5 + 2\sqrt{3}$ or $x = -5 - 2\sqrt{3}$

Solving $a(x - h)^2 = k$

When there is a coefficient in front of the squared binomial, divide it out first.

Example 8: $2(x - 1)^2 = 50$

Step 1 — Divide both sides by 2:

$(x - 1)^2 = 25$

Step 2 — Take the square root:

$x - 1 = \pm 5$

Step 3 — Solve both cases:

$x = 1 + 5 = 6 \quad \text{or} \quad x = 1 - 5 = -4$

Answer: $x = 6$ or $x = -4$

Example 9: $4(x + 2)^2 - 36 = 0$

Step 1 — Add 36 to both sides:

$4(x + 2)^2 = 36$

Step 2 — Divide by 4:

$(x + 2)^2 = 9$

Step 3 — Take the square root:

$x + 2 = \pm 3$

Step 4 — Solve:

$x = -2 + 3 = 1 \quad \text{or} \quad x = -2 - 3 = -5$

Answer: $x = 1$ or $x = -5$

Real-World Application: Electrician — Diagonal Conduit Run

An electrician needs to run conduit diagonally across a room corner where two walls meet at a right angle. The horizontal run along one wall is 18 inches and the vertical run along the other wall is 18 inches. The diagonal conduit length $L$ satisfies:

$L^2 = 18^2 + 18^2$

$L^2 = 324 + 324 = 648$

$L = \sqrt{648} = \sqrt{324 \cdot 2} = 18\sqrt{2}$

Since length must be positive, we take only the positive root.

$L \approx 25.46 \text{ inches}$

Answer: The diagonal conduit run is approximately 25.5 inches. The square root method gives the precise geometric length, and the electrician adds a few extra inches for fittings.

A Carpentry Application: Finding a Missing Measurement

A carpenter is building a square platform with an area of 72 square feet. What is the side length?

$s^2 = 72$

$s = \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} \approx 8.49 \text{ feet}$

Since a physical length is positive, we use only the positive root.

Answer: Each side is $6\sqrt{2} \approx 8$ feet $5\frac{7}{8}$ inches. The carpenter would likely round to 8 feet 6 inches for practical cutting.

Common Mistakes to Avoid

1. Forgetting the $\pm$ symbol. This is the most frequent error. $x^2 = 25$ gives $x = 5$ and $x = -5$, not just $x = 5$. Always write $\pm$ unless the context (like a physical length) eliminates the negative answer.

2. Taking square roots before isolating the squared term. You must get the squared expression alone on one side first. In $3x^2 - 12 = 0$, add 12 and divide by 3 before taking the square root.

3. Trying this method when there is a linear $x$ term. The equation $x^2 + 4x = 9$ cannot be solved by simply taking the square root of both sides. You would need to complete the square or use the quadratic formula.

4. Simplifying radicals incorrectly. Remember: $\sqrt{50} = 5\sqrt{2}$, not $25\sqrt{2}$. Factor out perfect squares from under the radical.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The square root method works when the variable appears only inside a squared expression — no separate linear $x$ term
• Always include $\pm$ when taking the square root of both sides — squaring hides the sign, and you must restore both possibilities
• Isolate the squared expression before taking the square root: divide out coefficients and move constants to the other side first
• When $x^2 = k$ and $k$ is negative, there are no real solutions
• Simplify radicals by factoring out perfect squares: $\sqrt{48} = 4\sqrt{3}$, not left as $\sqrt{48}$
• In real-world problems (lengths, distances), discard the negative root since physical measurements are positive

Return to Algebra 1 for more topics in this section.