The Square Root Property: Videos & Practice Problems

Using The Square Root Property

When solving quadratic equations, not all can be factored easily, so alternative methods are necessary. One effective technique is the square root property, which applies when a squared term is isolated on one side of the equation. If you have an equation in the form \(x^2 = k\), where \(k\) is a constant, you can solve for \(x\) by taking the square root of both sides. This process cancels the square and leaves you with \(x = \pm \sqrt{k}\). The plus-minus symbol (\(\pm\)) is crucial because both the positive and negative roots satisfy the equation, as squaring either will return the original positive value.

For example, if \(x^2 = 16\), then \(x = \pm \sqrt{16} = \pm 4\). Both \(4\) and \(-4\) are solutions since \(4^2 = 16\) and \((-4)^2 = 16\).

Consider the equation \(4x^2 - 8 = 0\). To solve using the square root property, first isolate the squared term by adding 8 to both sides: \(4x^2 = 8\). Then divide both sides by 4 to get \(x^2 = 2\). Applying the square root property gives \(x = \pm \sqrt{2}\). Since \(\sqrt{2}\) is already in simplest radical form, these are the final solutions.

Another example is when the squared term is a binomial, such as \((x + 1)^2 = 4\). Even though the squared quantity is more complex, the square root property still applies. Taking the square root of both sides yields \(x + 1 = \pm \sqrt{4} = \pm 2\). Solving for \(x\) involves subtracting 1 from both sides, resulting in \(x = -1 \pm 2\). This simplifies to two solutions: \(x = 1\) and \(x = -3\).

It is important to verify solutions by substituting them back into the original equation to ensure they satisfy it. The square root property is especially useful when the quadratic equation lacks a linear term (the \(bx\) term), or when the squared term is already isolated, even if it is a binomial. This method provides a straightforward way to solve such quadratics without factoring or using the quadratic formula.

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

• A. $y=\(\pm\)9$
• B. $y=\(\pm\)3\(\sqrt\)3$
• C. $y=27$
• D. $y=\(\pm\]\sqrt\)3$

Answer: (B)

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

• A. $x=\(\pm\)2$
• B. $x\(\pm\)3\(\sqrt\)2$
• C. $x=\(\pm\)16$
• D. $x=\(\pm\)4$

Answer: (D)

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

• A. $z=\(\pm\]\frac\)72$
• B. $z=-\(\frac\)12$ and $z=\(\frac\)72$
• C. $z=-\(\frac\)12$ and $z=\(\frac\)27$
• D. $z=-\(\frac\)27$ and $z=\(\frac\)72$

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 under gravity:

\(h = 9.8 t^2\),

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 h equal to 58 meters and solve for t. This gives the equation:

\(58 = 9.8 t^2\)

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

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

Taking the square root of both sides, we apply the square root property:

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

Since time cannot be negative in this context, we consider only the positive root. Calculating the value gives:

\(t \approx 2.4\) seconds

This means the drone takes approximately 2.4 seconds to fall from a height of 58 meters to the ground. This calculation demonstrates how the relationship between height and time in free fall is governed by the acceleration due to gravity, approximately 9.8 meters per second squared, and how solving quadratic equations can be applied to real-world physics problems.

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 understanding the relationship between the side length and the area. Since the garden is square-shaped, the area 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 solution 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. This process highlights the importance of understanding geometric formulas and applying algebraic techniques such as square roots to solve for unknown dimensions in real-world problems.

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 vertical 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 with legs a and b, and hypotenuse c, the relationship is:

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

Here, the height corresponds to a = 6 feet, and the hypotenuse is c = 19 feet. Substituting these 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 rewrite the square root as:

\[b = \sqrt{25 \times 13} = \sqrt{25} \times \sqrt{13} = 5 \sqrt{13}\]

Therefore, the base length of the ramp is:

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

This result demonstrates how the Pythagorean theorem can be applied to real-world problems involving right triangles, allowing for the calculation of unknown side lengths when two sides are known. The use of radical simplification ensures the answer is expressed in its simplest exact form, which is especially useful in geometry and construction contexts.

Imaginary Solutions

When solving quadratic equations using the square root property, encountering a negative number under the square root indicates the presence of imaginary or complex solutions. This occurs because the square root of a negative number is not defined within the real numbers but can be expressed using the imaginary unit i, where i is defined as the square root of -1, i.e., \(i = \sqrt{-1}\).

Consider the quadratic equation \(2x^2 + 32 = 0\(. To solve for x, first isolate the squared term by subtracting 32 from both sides:

\[2x^2 = -32\]

Next, divide both sides by 2 to simplify:

\[x^2 = -16\]

Applying the square root property involves taking the square root of both sides:

\[x = \pm \sqrt{-16}\]

Since the radicand (the number under the square root) is negative, rewrite the square root of -16 using the product property of square roots:

\[\sqrt{-16} = \sqrt{16} \times \sqrt{-1}\]

Recognizing that \)\sqrt{16} = 4\) and \(\sqrt{-1} = i\), the expression simplifies to:

\[x = \pm 4i\]

This shows that the solutions are purely imaginary numbers, \$4i\( and \)-4i$. Understanding how to manipulate and simplify expressions involving imaginary numbers is essential when solving quadratic equations that yield complex solutions. This approach extends the real number system to the complex number system, allowing for a complete set of solutions to all quadratic equations.

Solve each quadratic equation. If roots are not real, use $i$.
$2x^2+18=0$

• A. $x=\(\pm\)3$
• B. $x=\(\pm\]\frac\)92$
• C. $x=\(\pm\)3i$
• D. $x=\(\pm\]\frac\)92i$

Answer: (C)

Solve each quadratic equation. If roots are not real, use $i$.
$\frac{w^2}{4}+8=0$

• A. $w=\(\pm\)4\(\sqrt\)2i$
• B. $w=\(\pm\)4i$
• C. $w=\(\pm\)4$
• D. $w=\(\pm\)4\(\sqrt\)2$

Answer: (A)

Solve each quadratic equation. If roots are not real, use $i$.
${(3z-1)}^2+5=0$

• A. $z=\(\frac{1\pm2i}{3}\)$
• B. $z=\(\frac{1\pm\sqrt5i}{3}\)$
• C. $z=\(\pm\)3i$
• D. $z=\(\pm\]\frac\)13i$

Answer: (B)

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. It states that if $x^2=k$, then $x=\pm \sqrt{k}$. This means you take the square root of both sides, remembering to include both the positive and negative roots because squaring either a positive or negative number results in the same positive value. This method is especially useful when the quadratic equation lacks a linear term or is expressed as a binomial squared. For example, to solve $x^2=16$, you take the square root of both sides to get $x=\pm 4$. This property simplifies solving certain quadratics without factoring or using the quadratic formula.

How do you solve quadratic equations using the square root property when the value under the square root is negative?

When the value under the square root is negative, the solutions to the quadratic equation are imaginary or complex. For example, if you have $x^2=-16$, you apply the square root property by taking the square root of both sides: $x=\pm \sqrt{-16}$. Since the square root of a negative number is not real, you rewrite it using the imaginary unit $i$, where $\sqrt{-1}=i$. So, $\sqrt{-16}=\sqrt{16}\sqrt{-1}=\; 4i$. Therefore, the solutions are $x=\pm 4i$. This approach allows you to handle quadratics with no real solutions by expressing answers in terms of imaginary numbers.

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 and equals a constant. This typically happens when the quadratic lacks the linear term (the $bx$ term) or when the quadratic is expressed as a binomial squared, such as $x^2=k$ or ${x+1}^2=k$. In these cases, you can isolate the squared term and then take the square root of both sides, remembering to include both positive and negative roots. This method is not suitable for quadratics with a linear term unless you first complete the square to rewrite the equation in a suitable form.

How do you simplify solutions after applying the square root property to a binomial squared equation?

When you apply the square root property to an equation like ${x+1}^2=4$, you first take the square root of both sides to get $x+1=\pm 2$. Then, solve for $x$ by subtracting 1 from both sides: $x=\; -1\; \pm \; 2$. This means $x$ can be either $-1\; +\; 2\; =\; 1$ or $-1\; -\; 2\; =\; -3$. So, the solutions are $x=\; 1$ and $x=\; -3$. Simplifying in this way gives you the exact values of $x$ rather than leaving the answer in an unsimplified form with the plus-minus symbol.

How do you check your solutions after solving a quadratic equation using the square root property?

To check your solutions, substitute each solution back into the original quadratic equation to verify that it satisfies the equation. For example, if you solved $x^2=\; 16$ and found $x=\; \pm 4$, plug in $4$ and $-4$ separately. For $x\; =\; 4$, calculate $4^2\; =\; 16$, which matches the original equation. For $x\; =\; -4$, calculate $(-4)^2\; =\; 16$, also matching the original. If both satisfy the equation, your solutions are correct. This step is important to confirm accuracy, especially when dealing with complex or imaginary solutions.