The Square Root Property: Videos & Practice Problems

Solving quadratic equations often involves isolating a squared term and applying the square root property, which states that if $x^2=k$, then $x=\pm \sqrt{k}$. This method works even with binomials squared, such as ${x+1}^2$. When negative values appear under the square root, solutions involve imaginary numbers, using $i$ where $\sqrt{-1}=i$. Understanding coefficients, terms, and polynomial degrees aids in applying these concepts effectively.

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. $y=\pm9$
• 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 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. $x=\pm3$
• B. $x=\pm\frac92$
• C. $x=\pm3i$
• D. $x=\pm\frac92i$

Answer: (C)

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

• A. $w=\pm4\sqrt2i$
• B. $w=\pm4i$
• C. $w=\pm4$
• D. $w=\pm4\sqrt2$

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. $z=\pm3i$
• D. $z=\pm\frac13i$

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, considering both the positive and negative roots. It is especially useful when the quadratic equation lacks the linear term (the $bx$ term) or when dealing with binomials squared. For example, to solve $x^2=16$, you take the square root of both sides to get $x=\pm 4$. This method simplifies solving certain quadratics without factoring or using the quadratic formula.

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

When the number under the square root is negative, the solutions to the quadratic equation are imaginary or complex. Using the square root property, if you have $x^2=-k$ where $k$ is positive, you rewrite the square root of the negative number as the product of the square root of the positive number and the imaginary unit $i$, where $\sqrt{-1}=i$. For example, solving $2x^2+\; 32\; =\; 0$ leads to $x^2=\; -16$. Taking the square root gives $x=\; \pm \sqrt{-16}$, which simplifies to $x=\; \pm \; 4i$. This approach allows you to handle complex solutions effectively.

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 has the squared term isolated and either lacks the linear term ($bx$) or is in a form where a binomial is squared. For example, equations like $4x^2-\; 8\; =\; 0$ or $(x\; +\; 1)^2\; =\; 4$ are ideal for this method. Factoring may not be possible or efficient if the quadratic is not factorable with integers. The square root property simplifies the process by isolating the squared term and taking square roots, providing solutions quickly without the need for complex factoring or the quadratic formula.

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

When applying the square root property to binomial squares 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 isolating it: $x\; =\; -1\; \pm \; 2$. This gives two solutions: $x\; =\; 1$ and $x\; =\; -3$. Simplifying involves performing the addition and subtraction to find the exact values of $x$. This step is important because the ± symbol indicates two possible values, and simplifying clarifies the distinct solutions to the quadratic equation.

Can the square root property be used for all quadratic equations?

No, the square root property cannot be used for all quadratic equations. It is only applicable when the squared term is isolated on one side of the equation, such as $x^2=\; k$ or when a binomial is squared and isolated, like $(x\; +\; a)^2\; =\; k$. If the quadratic equation includes a linear term and cannot be easily rearranged to isolate the squared term, other methods like factoring, completing the square, or the quadratic formula are more appropriate. The square root property is a quick and effective method but limited to specific forms of quadratic equations.