Solving Quadratic Equations with the Square Root Property

Some quadratic equations can be solved without factoring. When the variable expression is already squared and isolated, use the square root property: if u² = k, then u = ±√k for real k ≥ 0. The ± symbol matters because a positive number and its negative have the same square. Applying only the principal, nonnegative square root would discard a valid solution.

The basic pattern

For (x − h)² = k, replace the squared expression with two cases: x − h = √k or x − h = −√k. Then add h to both sides. The solutions are x = h ± √k. For example, (x − 3)² = 25 becomes x − 3 = ±5, so x = 8 or x = −2. Check them in the original equation: both make the squared expression equal 25.

The shift inside the parentheses affects the final step. With (x + 4)² = 9, write x + 4 = ±3. The two cases give x = −1 and x = −7. A reliable habit is to keep the equation in the form x + 4 = ±3 until the last line, so the sign of the shift does not get lost.

Isolate the square first

The property applies after the squared expression is by itself. In 2(x − 1)² + 5 = 23, subtract 5 to get 2(x − 1)² = 18, divide by 2 to get (x − 1)² = 9, then take both roots: x − 1 = ±3. Thus x = 4 or x = −2. Each operation must be applied to both sides and preserve equivalence.

If the coefficient on the squared expression is negative, divide by that negative coefficient after isolating other terms. For −3(x + 2)² + 7 = −5, subtract 7 to obtain −3(x + 2)² = −12, then divide by −3: (x + 2)² = 4. The solutions are x = 0 and x = −4. Dividing by a negative changes inequality direction only in inequalities; an equation's equality sign does not reverse.

What if the right side is zero or negative?

If (x − h)² = 0, then x − h = 0 and there is one real solution, x = h. A square is never negative for a real number. Therefore, an equation such as (x + 1)² = −4 has no real solution. Some algebra courses extend numbers to complex values, but a real-number test question expects you to recognize that no real x can produce a negative square.

When the expression is not already a perfect square

The square root property still works when the right side is not a perfect square. If (x − 5)² = 7, then x = 5 ± √7. These are exact solutions; approximate decimals are optional unless requested. Keep the ± outside the radical: 5 + √7 and 5 − √7 are the two values.

A quadratic in the form x² = k is a special case with h = 0. For x² = 12, x = ±√12 = ±2√3. Simplify the radical if useful, but do not change the two-case structure. For 4x² = 20, first divide by 4 to get x² = 5, then x = ±√5. Taking the square root before isolating x² can lead to an incorrect coefficient.

Recognize when another method is needed

An equation such as x² + 6x = 7 is not ready for this method because the squared term and linear term are mixed. You could first rewrite it as x² + 6x − 7 = 0 and factor, or complete the square to create a squared binomial. Do not take the square root of each term separately: √(a + b) is not generally √a + √b. The method is direct only when you can isolate a single squared expression.

Common errors

• Forgetting the negative root and reporting only one solution when k is positive.
• Writing ± before isolating the squared expression.
• Taking the square root of a sum term by term.
• Treating a negative right side as if it had a real square root.
• Losing the sign when undoing a parenthesis such as x + 4.
• Failing to substitute both answers into the original equation.

A substitution check catches sign errors. For (x + 4)² = 9, x = −1 gives 3² = 9, and x = −7 gives (−3)² = 9. If you found only x = −1, the check of that value would not reveal the missing root; the ± step is essential, not optional.

Isolate the squared expression first

The square-root property applies after a squared quantity is isolated. From (x − 3)² = 16, take square roots of both sides: x − 3 = ±4. Then x = 3 ± 4, giving x = 7 or x = −1. The ± is essential because both 4² and (−4)² equal 16. Check both solutions in the original equation.

If the squared quantity equals a negative number, such as (x + 2)² = −9, there is no real solution because a real square cannot be negative. If complex numbers are in scope, solutions involve i; follow the problem’s domain. If the right side is zero, the positive and negative roots coincide at zero.

Exam takeaway

Isolate one squared expression, take the positive and negative square roots, and solve each case. A positive right side produces two real roots, zero produces one, and a negative right side produces none in the real numbers. Preserve the ± until you have solved for the variable.

Apply it to the evidence or sentence

The square root property applies once the squared expression is isolated: (x − 4)² = 25 gives x − 4 = ±5, so x = 9 or x = −1. Taking only the principal square root, 5, would omit a valid solution because both 5² and (−5)² equal 25. If the right side is zero, there is one repeated solution; if it is negative and the variables are restricted to real numbers, there is no real solution. When a coefficient appears, divide both sides before taking roots. Check each candidate in the original equation, especially after squaring or simplifying.