Algebra I/Quadratic Systems/Solving Quadratics by Taking Square Roots

Algebra I Regents in 22 days

Algebra I · Lesson 8

Solving Quadratics by Taking Square Roots

When a quadratic has no x term, you don't need to factor — just undo the square.

A skydiver jumps from a plane and falls freely for several seconds before opening a parachute. Physicists calculate how far an object falls using the equation $d = 16t^2$ , where $d$ is distance in feet and $t$ is time in seconds. If the skydiver falls 400 feet before pulling the cord, they solve $16t^2 = 400$ to find the time. There is no $t$ term — only a $t^2$ term. That structure lets you solve the equation by taking a square root instead of factoring.

The core idea is this: a square and a square root are inverse operations, the same way multiplication and division undo each other. If $x^2 = k$ , then $x = \pm\sqrt$ . The $\pm$ symbol is essential — both the positive and negative square roots satisfy the original equation, because both KaTeX can only parse string typed expression and KaTeX can only parse string typed expression equal $k$ .

x^2 = k \implies x = \pm\sqrt

Start with a clean example. Solve $x^2 = 49$ .

Solving x² = 49

x^2 = 49

The equation is already isolated — the squared term is alone on one side.

x = \pm\sqrt{49}

Take the square root of both sides. Write ± to capture both solutions.

x = \pm 7 \checkmark

7 squared is 49, and so is (-7) squared. Both values work.

Now try one where the answer is not a perfect square. Solve $x^2 = 18$ .

Solving x² = 18, simplified radical form

x^2 = 18

Take the square root of both sides.

x = \pm\sqrt{18}

18 is not a perfect square, so simplify the radical.

x = \pm\sqrt{9 \cdot 2}

Find the largest perfect square factor of 18 — that is 9.

x = \pm 3\sqrt{2} \checkmark

Pull the square root of 9 out front. The √2 stays inside because 2 has no perfect square factors.

The Regents will ask you to leave answers in simplified radical form. A radical is simplified when no perfect square remains inside the square root sign. Check by asking: does any factor appear twice under the radical? If yes, one of those factors comes outside.

The method extends naturally to equations where a binomial is squared. When the equation looks like $(x + a)^2 = k$ , treat the entire binomial as a single unit. Take the square root of both sides first, then solve for $x$ .

(x + a)^2 = k \implies x + a = \pm\sqrt, \implies x = -a \pm\sqrt

Solve $(x + 3)^2 = 25$ .

Solving (x + 3)² = 25

(x + 3)^2 = 25

The squared binomial is isolated — take the square root of both sides now.

x + 3 = \pm\sqrt{25}

Both sides get the square root. The ± goes on the right side.

x + 3 = \pm 5

√25 = 5, a perfect square. Now split into two separate equations.

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

The ± means there are two cases. Solve each one.

x = 2 \quad \text{or} \quad x = -8 \checkmark

Subtract 3 from both sides in each case.

Now a harder one with a radical answer. Solve $(x - 1)^2 = 12$ .

Solving (x − 1)² = 12

(x - 1)^2 = 12

Take the square root of both sides.

x - 1 = \pm\sqrt{12}

12 is not a perfect square — simplify before finishing.

x - 1 = \pm\sqrt{4 \cdot 3}

4 is the largest perfect square factor of 12.

x - 1 = \pm 2\sqrt{3}

Pull √4 = 2 out front. The √3 stays inside.

x = 1 \pm 2\sqrt{3} \checkmark

Add 1 to both sides. This is the final simplified form — two irrational solutions.

The two solutions here are $x = 1 + 2\sqrt$ and $x = 1 - 2\sqrt$ . Both are exact. A decimal approximation would lose precision — the Regents expects the exact form unless the problem says otherwise.

Here is a visual of these two ideas. The graph of $y = x^2$ and the horizontal line $y = k$ intersect at exactly the two solutions $x = \pm\sqrt$ when $k > 0$ . When $k = 0$ , there is one solution. When $k < 0$ , the parabola never reaches the line and there are no real solutions.

Interactive graph — scroll to zoom, drag to pan

Solve each equation. Simplify all radical answers.

Practice Questions

x^2 = 64

x^2 = 50

(x + 5)^2 = 9

(x - 4)^2 = 20

Regents Corner

On Part II and Part III of the Algebra I Regents, solving by square roots appears both in isolation and as a step inside longer problems. A complete answer requires both solutions — writing only the positive root loses a point. Show the $\pm$ symbol explicitly at the step where you take the square root, and keep it through the rest of the work so the grader can follow your reasoning.

Students solve (x + 3)² = 25 and write x = ±5 − 3, which gives x = 2 or x = −8 correctly by luck of arithmetic — but then on a problem like (x + 3)² = 7 they write x = ±√7 − 3 instead of x = −3 ± √7. These are the same values, but writing −3 ± √7 matches the expected form and makes the two solutions unmistakably clear. Subtract or add the constant after the ± to avoid ambiguity.

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Solving Quadratics by Factoring

The Discriminant and Nature of Roots