Unit 9 · Lesson 1 of 5

Solving quadratics with the square root property

About 20 minutes · 5 steps · one idea: undo a square with ±√

You'll be able to: solve equations like \(x^2 = k\) and \(a(x-h)^2 = k\), including ones whose solutions are radicals or complex numbers.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.

Write \(\sqrt{-18}\) in simplest form using \(i\).

\(\sqrt{-18} = i\sqrt{18}\). Then \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\).

Answer: \(3i\sqrt{2}\) (you can type 3sqrt(2)i).

Solve by factoring: \(x^2 - 2x - 15 = 0\).

Two numbers that multiply to \(-15\) and add to \(-2\): \(-5\) and \(3\). So \((x - 5)(x + 3) = 0\).

Set each factor to 0: \(x = 5\) or \(x = -3\).

Step 2 of 5The idea

Some quadratic equations have only a squared part and a number, like \(x^2 = 25\) or \((x - 3)^2 = 12\). For these you don't need to factor. You can undo the square with a square root.

The catch: \(5^2 = 25\) and also \((-5)^2 = 25\). So \(x^2 = 25\) has two solutions, \(5\) and \(-5\). We write both at once as \(x = \pm 5\), read "plus or minus 5".

Square root property. If \(x^2 = k\), then \(x = \pm\sqrt{k}\). The same works for any squared part: if \((x - h)^2 = k\), then \(x - h = \pm\sqrt{k}\).

Steps. 1. Get the squared part alone. 2. Take the square root of both sides and write ±. 3. Simplify the radical. 4. Finish solving for \(x\).

Negative k. The square root of a negative uses \(i\): \(\sqrt{-k} = i\sqrt{k}\). The solutions are complex numbers.

This is the square root property. Always write answers as simplified radicals, like \(2\sqrt{3}\) instead of \(\sqrt{12}\).

Why this shows up on the placement test: equations like \((x+2)^2 = 18\) are common test items, and this property is the last step of completing the square in the next lesson.

Step 3 of 5Worked examples

Example 1: \(3x^2 - 7 = 41\)

  1. Add 7 to both sides: \(3x^2 = 48\).
  2. Divide both sides by 3: \(x^2 = 16\). Now the squared part is alone.
  3. Take the square root of both sides, with ±: \(x = \pm\sqrt{16} = \pm 4\).
  4. The solutions are \(x = 4\) and \(x = -4\). Check: \(3(16) - 7 = 41\), and \((-4)^2\) is also 16.

Example 2: \(2(x - 3)^2 = 24\), then \((x + 1)^2 + 9 = 0\)

  1. Divide both sides by 2: \((x - 3)^2 = 12\). The squared part is \((x - 3)\). Keep it in one piece; don't multiply it out.
  2. Root with ±: \(x - 3 = \pm\sqrt{12}\).
  3. Simplify: \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\), so \(x - 3 = \pm 2\sqrt{3}\).
  4. Add 3 to both sides: \(x = 3 \pm 2\sqrt{3}\). Two solutions: \(3 + 2\sqrt{3}\) and \(3 - 2\sqrt{3}\).
  5. Second equation. Subtract 9: \((x + 1)^2 = -9\).
  6. Root with ±: \(x + 1 = \pm\sqrt{-9} = \pm 3i\). \(\sqrt{-9} = i\sqrt{9} = 3i\).
  7. Subtract 1: \(x = -1 \pm 3i\).

Step 4 of 5Practice

Type your answer and press Check. List every solution, separated by commas, or use ± (type +- or ±). If you miss twice, open the worked solution. Then mark how it went.

Solve \(x^2 = 49\).

Take the square root of both sides, with ±: \(x = \pm\sqrt{49} = \pm 7\).

Answer: \(x = 7\) or \(x = -7\).

Solve \(x^2 = 20\). Simplify the radical.

\(x = \pm\sqrt{20}\).

\(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\).

Answer: \(x = \pm 2\sqrt{5}\).

Solve \(4x^2 - 9 = 0\).

Add 9: \(4x^2 = 9\). Divide by 4: \(x^2 = \dfrac{9}{4}\).

Root with ±: \(x = \pm\sqrt{\dfrac{9}{4}} = \pm\dfrac{3}{2}\).

Answer: \(x = \dfrac{3}{2}\) or \(x = -\dfrac{3}{2}\).

Solve \((x - 5)^2 = 36\).

Root with ±: \(x - 5 = \pm 6\).

Add 5: \(x = 5 + 6 = 11\) or \(x = 5 - 6 = -1\).

Answer: \(11, -1\).

Solve \((x + 2)^2 = 18\).

Root with ±: \(x + 2 = \pm\sqrt{18}\).

Simplify: \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\), so \(x + 2 = \pm 3\sqrt{2}\).

Subtract 2: \(x = -2 \pm 3\sqrt{2}\).

Solve \(x^2 + 16 = 0\).

Subtract 16: \(x^2 = -16\).

Root with ±: \(x = \pm\sqrt{-16} = \pm i\sqrt{16} = \pm 4i\).

Answer: \(x = 4i\) or \(x = -4i\).

Solve \(3(x - 1)^2 + 15 = 0\).

Subtract 15: \(3(x - 1)^2 = -15\). Divide by 3: \((x - 1)^2 = -5\).

Root with ±: \(x - 1 = \pm\sqrt{-5} = \pm i\sqrt{5}\).

Add 1: \(x = 1 \pm i\sqrt{5}\).

Which gives all the solutions of \((2x - 1)^2 = 8\)?

  1. \(x = \dfrac{1 \pm 4\sqrt{2}}{2}\)
  2. \(x = \dfrac{1 + 2\sqrt{2}}{2}\)
  3. \(x = \dfrac{-1 \pm 2\sqrt{2}}{2}\)
  4. \(x = \dfrac{1 \pm 2\sqrt{2}}{2}\)

Root with ±: \(2x - 1 = \pm\sqrt{8} = \pm 2\sqrt{2}\), because \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\).

Add 1: \(2x = 1 \pm 2\sqrt{2}\). Divide by 2: \(x = \dfrac{1 \pm 2\sqrt{2}}{2}\). That is (d).

(a) simplifies \(\sqrt{8}\) wrongly as \(4\sqrt{2}\). (b) forgets the ±, so it misses one solution. (c) moves the 1 with the wrong sign.

Step 5 of 5Wrap-up

You can now solve any equation that is "a square equals a number": get the square alone, take ±√ of both sides, simplify the radical (using \(i\) for negatives), and finish solving.

Read or watch next: OpenStax Intermediate Algebra 2e, 9.1 Solve Quadratic Equations Using the Square Root Property (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Solving quadratics by taking square roots (short videos). Quick rules: Quadratics cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why can't I take the square root of each term in \(x^2 + 9 = 25\) separately?" It knows this course and can explain things another way.