The Square Root Property: Videos & Practice Problems

Solving quadratic equations often involves factoring, but when a polynomial lacks a linear term or is in a binomial squared form, the square root property is essential. This method applies when an exponent of 2 isolates a base, allowing us to take the square root of both sides, yielding $x=\pm \sqrt{k}$. Recognizing coefficients and terms in standard form helps simplify expressions, especially with perfect square trinomials. This approach deepens understanding of polynomials, exponents, and solving techniques beyond factoring, crucial for mastering quadratic equations and their solutions.

Using The Square Root Property

When solving quadratic equations, not all can be factored easily, so alternative methods like the square root property become essential. The square root property is particularly useful when a squared term is isolated on one side of the equation, such as in expressions of the form x² = k, where k is a constant. To solve for x, you take the square root of both sides, which cancels the square on the left, leaving x = ±√k. The ± symbol indicates both the positive and negative roots because squaring either a positive or negative number results in the same positive value.

For example, if x² = 16, applying the square root property gives x = ±√16, which simplifies to x = ±4. Both 4 and -4 satisfy the original equation since 4² = 16 and (-4)² = 16.

Consider the equation 4x² - 8 = 0. To solve using the square root property, first isolate the squared term by adding 8 to both sides, resulting in 4x² = 8. Then divide both sides by 4 to get x² = 2. Applying the square root property yields x = ±√2. Since √2 is an irrational number and cannot be simplified further, these are the final solutions.

In cases where the squared term is a binomial, such as (x + 1)² = 4, the square root property still applies. Taking the square root of both sides gives x + 1 = ±2. Solving for x involves subtracting 1 from both sides, resulting in x = -1 ± 2. This simplifies to two solutions: x = 1 and x = -3.

The square root property is most effective when the quadratic equation lacks the linear term (the b coefficient is zero) or when the equation is already in a form where a binomial is squared. Always verify solutions by substituting them back into the original equation to ensure correctness.

In summary, the square root property provides a straightforward method to solve quadratic equations when the squared term is isolated. By taking the square root of both sides and considering both positive and negative roots, you can find all possible solutions efficiently.

Solve each quadratic equation using the square root property.
$2y^2=54$

• A. $y=\pm9$
• B. $y=\pm3\sqrt3$
• C. $y=27$
• D. $y=\pm\sqrt3$

Answer: (B)

Solve each quadratic equation using the square root property.
$3x^2+3=51$

• A. $x=\pm2$
• B. $x\pm3\sqrt2$
• C. $x=\pm16$
• D. $x=\pm4$

Answer: (D)

Solve each quadratic equation using the square root property.
${(2z-3)}^2=16$

• A. $z=\pm\frac72$
• B. $z=-\frac12$ and $z=\frac72$
• C. $z=-\frac12$ and $z=\frac27$
• D. $z=-\frac27$ and $z=\frac72$

Answer: (B)

Using The Square Root Property Example 1

When a drone is dropped from a height of 58 meters, we can determine the time it takes to hit the ground using the equation for free fall motion:

\(h = 9.8 t^2\),

where h is the height in meters and t is the time in seconds. To find the time when the drone reaches the ground, we set the height equal to 58 meters and solve for t. Substituting, we get:

\$58 = 9.8 t^2\(

Dividing both sides by 9.8 to isolate \)t^2\( gives:

\)\frac{58}{9.8} = t^2\(

Applying the square root to both sides to solve for t results in:

\)t = \pm \sqrt{\frac{58}{9.8}}\(

Since time cannot be negative in this context, we only consider the positive root. Calculating the square root yields approximately:

\)t \approx 2.4$ seconds

This means the drone takes about 2.4 seconds to fall from a height of 58 meters to the ground. This calculation uses the acceleration due to gravity, approximately 9.8 meters per second squared, which is a key constant in free fall motion equations. Understanding how to manipulate and solve quadratic equations like this is essential for analyzing motion under constant acceleration.

Using The Square Root Property Example 2

To determine the length of one side of a square garden with an area of 196 square feet, we start by recognizing that the area of a square is calculated by squaring the length of one side. If we denote the side length as s, then the area A is given by the formula:

\[ A = s^2 \]

Given that the area is 196 square feet, we substitute this value into the equation:

\[ 196 = s^2 \]

To solve for s, we apply the square root property, which involves taking the square root of both sides:

\[ s = \pm \sqrt{196} \]

Since length cannot be negative in this context, we discard the negative root and keep only the positive value:

\[ s = \sqrt{196} \]

Calculating the square root of 196 yields:

\[ s = 14 \]

Therefore, the length of one side of the square garden is 14 feet.

Using The Square Root Property Example 3

To determine the length of the base of a ramp forming a right triangle with the ground, we use the Pythagorean theorem, which relates the lengths of the sides in a right triangle. Given that the ramp's hypotenuse measures 19 feet and the height from the ground to the platform is 6 feet, we want to find the base length along the ground, denoted as b.

The Pythagorean theorem states that for a right triangle, the sum of the squares of the two legs equals the square of the hypotenuse:

\[a^2 + b^2 = c^2\]

Here, c is the hypotenuse (19 feet), and a and b are the legs of the triangle, with a being the height (6 feet). Substituting the known values gives:

\[6^2 + b^2 = 19^2\]

Calculating the squares:

\[36 + b^2 = 361\]

Isolating b² by subtracting 36 from both sides:

\[b^2 = 361 - 36 = 325\]

To find b, take the square root of both sides:

\[b = \pm \sqrt{325}\]

Since length cannot be negative, we consider only the positive root. Next, simplify the radical by factoring 325:

\[325 = 25 \times 13\]

Because 25 is a perfect square, we can extract it from under the square root:

\[b = 5 \sqrt{13}\]

Therefore, the base of the ramp measures 5√13 feet. This solution demonstrates how the Pythagorean theorem can be applied to real-world problems involving right triangles, enabling the calculation of unknown side lengths when two sides are known.

Here’s what students ask on this topic:

What is the square root property and how is it used to solve quadratic equations?

The square root property is a method used to solve quadratic equations when the squared term is isolated on one side of the equation, such as $x^2=k$, where $k$ is a constant. To solve, you take the square root of both sides, which cancels the square on the left, giving $x=\; \pm \sqrt{k}$. The ± symbol indicates there are two solutions: one positive and one negative, because both a positive and negative number squared will result in the same positive value. This method is especially useful when the quadratic equation does not have a linear term (the $bx$ term) or when the quadratic is a perfect square trinomial. It simplifies solving quadratics that cannot be factored easily.

How do you apply the square root property when the squared term is a binomial, like (x + 1)² = 4?

When the squared term is a binomial, such as ${x+1}^2=4$, you can still apply the square root property. First, take the square root of both sides: $x+1=\; \pm \sqrt{4}$. This simplifies to $x+1=\; \pm 2$. Next, solve for $x$ by subtracting 1 from both sides, resulting in $x=\; -1\pm 2$. This gives two solutions: $x=\; 1$ and $x=\; -3$. This method works because the square root property applies to any quantity squared, whether a single variable or a binomial.

When is it appropriate to use the square root property instead of factoring to solve quadratic equations?

The square root property is appropriate when the quadratic equation is in the form $x^2=k$ or when a binomial squared equals a constant, and the equation does not have a linear term (the $bx$ term). This often happens when the coefficient of the linear term is zero, making factoring difficult or impossible. For example, equations like $4x^2-8=\; 0$ can be rearranged to isolate the squared term and then solved using the square root property. It is also useful for perfect square trinomials that can be rewritten as a binomial squared. If factoring is complicated or the quadratic is not factorable, the square root property provides a straightforward alternative.

How do you simplify the solutions after applying the square root property?

After applying the square root property, you get solutions in the form $x=\; \pm \sqrt{k}$. To simplify, first calculate the square root of $k$ if it is a perfect square. For example, $\sqrt{16}=4$, so the solutions are $x=\; \pm 4$. If $k$ is not a perfect square, leave the solution in radical form, such as $x=\; \pm \sqrt{2}$. If the squared term is a binomial, after taking the square root, isolate $x$ by performing inverse operations like addition or subtraction. Finally, express the solutions as two separate values by evaluating the positive and negative cases. This ensures the solutions are clear and fully simplified.

Can the square root property be used if the constant on the right side of the equation is negative?

If the constant on the right side of the equation is negative, such as $x^2=\; -9$, the square root property still applies, but the solutions will involve imaginary numbers because the square root of a negative number is not real. Taking the square root of both sides gives $x=\; \pm \sqrt{-9}$, which simplifies to $x=\; \pm 3i$, where $i$ is the imaginary unit. This means the solutions are complex numbers. So, while the square root property can be used, the nature of the solutions depends on the sign of the constant. For real solutions, the constant must be non-negative.