The Square Root Property: Videos & Practice Problems

When solving quadratic equations, the square root property is essential for equations where a base or binomial is squared and isolated, such as $x^2=k$. By taking the square root of both sides, including both positive and negative roots, students can find solutions even when factoring is not possible. This method applies to polynomials missing the linear term (coefficient of b is zero) or when dealing with perfect square trinomials. Understanding this property enhances problem-solving skills for polynomials and exponential expressions in standard form.

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 a term is squared and isolated on one side of the equation, such as $x^2=k$. To solve, you take the square root of both sides, which cancels the square on the variable, resulting in $x=\pm \sqrt{k}$. The ± symbol indicates there are two solutions: one positive and one negative, because both positive and negative values squared give the same positive result. This method is especially useful when the quadratic equation lacks a linear term (the $bx$ term) or when the equation is a perfect square trinomial. By isolating the squared term first, the square root property provides a straightforward way to find the roots of the quadratic.

When can the square root property be applied to solve quadratic equations?

The square root property can be applied when the quadratic equation has a squared term isolated on one side, such as $x^2=k$, where $k$ is a constant. This often occurs when the quadratic equation is missing the linear term ($bx$ term), meaning the coefficient $b$ is zero, or when the equation is written as a perfect square binomial, like ${x+1}^2=4$. In these cases, you can isolate the squared term and then take the square root of both sides to solve for the variable. This method is not suitable if the squared term is not isolated or if the equation is not easily rearranged into this form.

How do you solve an equation like (x + 1)² = 4 using the square root property?

To solve ${x+1}^2=4$ using the square root property, first take the square root of both sides: $x+1=\pm \sqrt{4}$. Since $\sqrt{4}=2$, this becomes $x+1=\pm 2$. Next, solve for $x$ by subtracting 1 from both sides: $x=\pm 2-1$. This gives two solutions: $x=1$ and $x=-3$. Both values satisfy the original equation when substituted back.

Why do we include both positive and negative roots when using the square root property?

When solving equations using the square root property, both positive and negative roots are included because squaring either a positive or negative number results in the same positive value. For example, $4^2=16$ and ${-}^4=16$. Therefore, if $x^2=16$, both $x=4$ and $x=-4$ are valid solutions. The ± symbol is used to represent this dual possibility, ensuring that all solutions to the quadratic equation are considered.

How do you solve an equation like 4x² - 8 = 0 using the square root property?

To solve $4x^2-8=0$, first isolate the squared term by adding 8 to both sides: $4x^2=8$. Next, divide both sides by 4: $x^2=2$. Now apply the square root property by taking the square root of both sides: $x=\pm \sqrt{2}$. Since the square root of 2 cannot be simplified further, the solutions are $x=\pm \sqrt{2}$. These represent the two values of $x$ that satisfy the original equation.