The Square Root Property: Videos & Practice Problems

Solving quadratic equations often involves isolating a squared term and applying the square root property, which yields solutions as both positive and negative roots. When the radicand is negative, solutions become imaginary, incorporating the imaginary unit $i$. This method applies to polynomials missing the linear term or expressed as binomials squared. Understanding coefficients, exponents, and polynomial structure aids in simplifying expressions and verifying solutions. Mastery of these concepts enhances problem-solving skills with monomials, binomials, and complex numbers in standard form, crucial for algebraic proficiency.

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. For an equation of 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 square root, leaving \(x = \pm \sqrt{k}\). The plus-minus symbol (\(\pm\)) indicates that both the positive and negative roots are solutions because squaring either a positive or negative number results in a positive value.

For example, if \(x^2 = 16\), then \(x = \pm \sqrt{16} = \pm 4\). Both \$4\( and \)-4\( satisfy the original equation since \)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}\( is an irrational number and cannot be simplified further, the solutions remain \)x = \pm \sqrt{2}\(.

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\(. This expression represents two solutions: \)x = -1 + 2 = 1\( and \)x = -1 - 2 = -3\(.

The square root property is particularly useful when the quadratic equation lacks the linear term (the \)bx\( term), meaning the coefficient \)b\( is zero, or when the equation is already in a form where a binomial is squared. Always remember to check your solutions by substituting them back into the original equation to verify their correctness.

In summary, the square root property provides a straightforward method to solve quadratic equations of the form \)x^2 = k\( or \)(x + c)^2 = k$ by isolating the squared term and taking the square root of both sides, considering both positive and negative roots. This method expands the toolkit for solving quadratics beyond factoring, especially when factoring is not feasible.

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. $z=\pm \frac{7}{2}$
• B. $z=-\frac{1}{2}$ and $z=\frac{7}{2}$
• C. $z=-\frac{1}{2}$ and $z=\frac{2}{7}$
• D. $z=-\frac{2}{7}$ and $z=\frac{7}{2}$

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 consider only 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 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:

\[36 + b^2 = 361\]

To isolate b², subtract 36 from both sides:

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

Next, apply the square root to both sides to solve for b:

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

Since length cannot be negative, we consider only the positive root. To simplify the radical, factor 325:

\[325 = 25 \times 13\]

Recognizing that 25 is a perfect square, we extract it from under the square root:

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

Therefore, the base length of the ramp is 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.

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, or \(i = \sqrt{-1}\).

For example, 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, take the square root of both sides:

\)x = \pm \sqrt{-16}\(

Since the radicand is negative, rewrite the square root using the product property:

\)\sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i\(

Thus, the solutions are:

\)x = \pm 4i$

This process demonstrates how to handle imaginary solutions by expressing the square root of a negative number in terms of the imaginary unit i. Understanding this allows for solving quadratic equations that do not have real roots, expanding the scope of solutions to include complex numbers. Mastery of this technique is essential for working with complex solutions in algebra and higher-level mathematics.

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

• A. $x=\pm 3$
• B. $x=\pm \frac{9}{2}$
• C. $x=\pm 3i$
• D. $x=\pm \frac{9}{2}i$

Answer: (C)

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

• A. $w=\pm 4\sqrt{2}i$
• 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. $z=\frac{1\pm 2i}{3}$
• B. $z=\frac{1\pm \sqrt{5}i}{3}$
• C. $z=\pm 3i$
• D. $z=\pm \frac{1}{3}i$

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 on one side of the equation. If you have an equation of the form $x^2=k$, where $k$ is a constant, you can solve for $x$ by taking the square root of both sides. This gives $x=\pm \sqrt{k}$, meaning there are two solutions: the positive and negative square roots of $k$. This works because both a positive and negative number squared will result in the same positive value. The square root property is especially useful when the quadratic equation does not have a linear term (the $bx$ term) or when the equation is expressed as a binomial squared.

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

When solving quadratic equations using the square root property, if the value under the square root (the radicand) is negative, the solutions are imaginary or complex numbers. For example, if you have $x^2=-16$, taking the square root of both sides gives $x=\pm \sqrt{-16}$. To simplify, rewrite the square root of the negative number as the product of the square root of the positive number and the square root of $-1$. Since $\sqrt{-1}=i$ (the imaginary unit), the solution becomes $x=\pm 4i$. This method allows you to handle quadratic equations 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 in two cases: first, when the quadratic is missing the linear term (the $bx$ term), so the equation looks like $a{x}^2=c$; second, when the quadratic is expressed as a binomial squared, such as ${x+1}^2=k$. In both cases, you isolate the squared term or binomial squared and then take the square root of both sides, remembering to include both positive and negative roots. This method is straightforward and efficient for these specific forms of quadratic equations.

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

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 there are two solutions: $x=1$ (from $-1+2$) and $x=-3$ (from $-1-2$). Simplifying in this way helps you find the exact numerical solutions rather than leaving the answer in terms of the ± symbol.

How do you verify solutions obtained using the square root property?

To verify solutions found using the square root property, substitute each solution back into the original quadratic equation. For example, if your solutions are $x=\pm 4$ for $x^2=16$, plug in $x=4$ and $x=-4$ separately. Calculate the left side of the equation for each value and check if it equals the right side. If both satisfy the equation, your solutions are correct. This step is important to confirm accuracy, especially when dealing with complex or imaginary solutions.