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Chapter 8: Problem 22

Use the square root property to solve each equation. \(x^{2}=54\)

Short Answer

Expert verified
x = \pm 3 \sqrt{6}

Step by step solution

01

- Isolate the variable term

The equation is already in the form where the variable term is isolated: \[ x^2 = 54 \]
02

- Apply the square root property

To solve for \( x \), take the square root of both sides of the equation:\[ x = \pm \sqrt{54} \]
03

- Simplify the square root

Simplify \( \sqrt{54} \): \[ \sqrt{54} = \sqrt{9 \times 6} = \sqrt{9} \times \sqrt{6} = 3 \sqrt{6} \]So, \[ x = \pm 3 \sqrt{6} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving equations
Solving equations is a fundamental part of algebra that you'll encounter often. The key goal is to find the values of the variables that make the equation true.

Here's a simple guide to solving equations:
  • Identify the variable you need to solve for.
  • Isolate the variable on one side of the equation. This often means using addition, subtraction, multiplication, or division to move other terms to the opposite side.
  • Simplify both sides of the equation as much as possible.
In our example, the equation given was \( x^2 = 54 \). The variable \( x \) was already isolated, making our task easier. We then used properties of algebra to find its value.
simplifying square roots
Simplifying square roots is an essential skill for making complicated expressions more manageable.

For example, simplifying \( \sqrt{54} \):
  • Factor the number under the square root to find its prime factors. Here, \( 54 = 9 \times 6 \).
  • Recognize perfect squares among these factors. In our case, \( 9 \) is a perfect square since \( 9 = 3^2 \).
  • Break down the square root into those factors. \( \sqrt{54} = \sqrt{9 \times 6} \).
  • Simplify further using properties of square roots: \( \sqrt{9} \times \sqrt{6} = 3\sqrt{6} \).
So, \( \sqrt{54} \) was simplified to \( 3\sqrt{6} \). This step makes the equation much easier to work with and understand.
quadratic equations
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). In our example, \( x^2 = 54 \), it's a simplified form of a quadratic equation where \( a = 1 \), \( b = 0 \), and \( c = -54 \).

Solving quadratic equations can often be done by the square root property, especially when the equation is as simple as our example.
  • Write the equation in standard form, if it's not already: \( x^2 - 54 = 0 \).
  • Use the square root property to solve for \( x \): \( x = \pm \sqrt{54} \).
  • Simplify your answer as needed: \( x = \pm 3 \sqrt{6} \).
This method is very efficient for equations where one term is squared and no linear term \( (bx) \) is present. Always simplify your radicals to get the cleanest form of your answer.

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Most popular questions from this chapter

The following exercises are not grouped by type. Solve each equation. $$8 x^{6}+513 x^{3}+64=0$$

Solve each inequality, and graph the solution set. $$ \frac{x-8}{x-4}<3 $$

In the 1939 classic movie The Wizard of Oz, Ray Bolger's character, the Scarecrow, wants a brain. When the Wizard grants him his "Th.D." (Doctor of Thinkology), the Scarecrow replies with the following statement. Scarecrow: The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. His statement sounds like the formula for the Pythagorean theorem. In the Scarecrow's statement, he refers to square roots. In applying the formula for the Pythagorean theorem, do we find square roots of the sides? If not, what do we find?

A ball is projected upward from ground level, and its distance in feet from the ground in t seconds is given by $$ s(t)=-16 t^{2}+160 t $$ After how many seconds does the ball reach a height of \(425 \mathrm{ft}\) ? Interpret the mathematical result here

Extending Skills Find the value of \(a, b\), or \(c\) so that each equation will have exactly one rational solution. One solution of \(4 x^{2}+b x-3=0\) is \(-\frac{5}{2} .\) Find \(b\) and the other solution.
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