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Chapter 8: Problem 19

Use the square root property to solve each equation. See Example 1. $$ z^{2}-50=0 $$

Short Answer

Expert verified
The solutions are \( z = 5\sqrt{2} \) and \( z = -5\sqrt{2} \).

Step by step solution

01

Rewrite the Equation in Standard Form

Start by rewriting the given equation in the standard form of a quadratic equation: \[ z^2 - 50 = 0 \] This equation is already in standard form, \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 0 \), and \( c = -50 \).
02

Isolate the Square Term

Add 50 to both sides of the equation to isolate the square term on one side of the equation:\[ z^2 = 50 \]
03

Apply the Square Root Property

Take the square root of both sides of the equation. Remember that taking the square root introduces both a positive and negative solution:\[ z = \pm \sqrt{50} \]
04

Simplify the Square Root

Simplify the expression \( \sqrt{50} \) into its simplest radical form by factoring under the square root:\[ \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \]
05

Write the Final Solutions

The solutions to the equation are:\[ z = 5\sqrt{2} \quad \text{or} \quad z = -5\sqrt{2} \]

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Key Concepts

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

Square Root Property
The square root property is a handy tool when solving quadratic equations, particularly when the equation is already set up in a simple form. This property is used to solve equations of the format \( x^2 = a \). The essence of the square root property is that you can solve for \( x \) by taking the square root of both sides of the equation. This process results in two possible solutions: a positive and a negative value.
  • Why Two Solutions? Because both \( x \) and \( -x \) squared will give you \( a \). Hence, if \( a = 50 \), then \( x = 5\sqrt{2} \) and \( x = -5\sqrt{2} \) are both valid solutions.
  • When to Use: Apply the square root property when your equation can be expressed in the form \( x^2 = a \) after some manipulation, like adding or subtracting numbers.
When utilizing this method, remember to simplify any radicals that arise, as part of achieving the simplest form for your solutions.
Simplifying Radicals
Simplifying radicals involves breaking down a square root into its simplest form. This conversion helps in effectively presenting the solution. To simplify a radical:
  • Factor the Number: Begin by factoring the number inside the square root into a product that includes a perfect square. For example, if you have \( \sqrt{50} \), you can express this as \( \sqrt{25 \times 2} \).
  • Square Root of Perfect Square: Next, find the square root of the perfect square; \( \sqrt{25} \) results in 5. So, \( \sqrt{50} \) simplifies to \( 5\sqrt{2} \).
  • Resulting Simplicity: This gives us a much cleaner expression, which in this case is \( 5\sqrt{2} \), making it easier to work with and interpret in further calculations.
Simplifying radicals is crucial since it leads to neater, reduced forms of your quadratic solutions, making them easier to understand and communicate.
Standard Form of Quadratic Equation
The standard form of a quadratic equation is written as \( ax^2 + bx + c = 0 \). It's a conventional way to express quadratic equations and acts as a critical step in solving them effectively.
  • Identify the Coefficients: Here, \( a \) represents the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \) is the constant. For the equation \( z^2 - 50 = 0 \), we have \( a = 1 \), \( b = 0 \), and \( c = -50 \).
  • Why It's Important: Having the equation in this format allows one to more easily apply solutions methods like the quadratic formula, completing the square, or factoring.
  • Simplification: Rearranging an equation into standard form can sometimes require simple operations such as moving terms across the equals sign and applying basic arithmetic. This step ensures that you have a clear path toward solving the equation.
By getting your quadratic equation into the standard form, you set the stage for simplifying and solving the equation accurately and concisely.

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Most popular questions from this chapter

a. \(a^{2}+4 a-7=0\) b. \(a^{2}-4 a-7=0\)

All of the equations we have solved so far have had rational-number coefficients. However, the quadratic formula can be used to solve quadratic equations with irrational or even imaginary coefficients. Solve each equation. $$ 100 i x^{2}+300 x-200 i=0 $$

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A table of values for \(f(x)=2 x^{2}-4 x+3\) is shown Explain why it appears that the vertex of the graph of \(f\) is the point \((1,1)\) (IMAGE CANNOT COPY)

All of the equations we have solved so far have had rational-number coefficients. However, the quadratic formula can be used to solve quadratic equations with irrational or even imaginary coefficients. Solve each equation. $$ x^{2}+2 \sqrt{2} x-6=0 $$
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