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Chapter 0: Problem 107

Solve each equation in Exercises \(107-120\) for \(x\), rounding your answer to four significant digits where necessary. $$ x^{2}-16=0 $$

Short Answer

Expert verified
The equation \(x^2-16=0\) has two possible solutions, \(x=4\) and \(x=-4\).

Step by step solution

01

Identify the constant term

The given quadratic equation is \(x^2 - 16 = 0\). Here, the constant term is -16.
02

Use the square root property to isolate x

Add 16 to both sides of the equation: \(x^2 - 16 + 16 = 0 + 16\) This simplifies to: \[x^2 = 16\] Now, use the square root property by taking the square root of both sides. Remember that when we take the square root of a variable squared, we must include both positive and negative solutions: \[\pm\sqrt{x^2} = \pm\sqrt{16}\]
03

Simplify and round if necessary

After taking the square root, we obtain the following solutions for 'x': \[\pm x = \pm 4\] Since the square root of 16 is a whole number (4), there is no need to round the answer. The two possible values for 'x' are: \[x = 4\] and \[x = -4\]

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Most popular questions from this chapter

\(\sqrt{(x+1)^{3}}+\sqrt{(x+1)^{5}}\)

\((x+1)(x+2)^{2}+(x+1)^{2}(x+2)\)

Find all possible real solutions of each equation \(\left(x^{2}+3 x+2\right)\left(x^{2}-5 x+6\right)=0\)

Convert each expression in Exercises into its technology formula equivalent as in the table in the text.\(4+\frac{5}{9}\)

By any method, determine all possible real solutions of each equation Check your answers by substitution. \(2 x^{2}-4 x+3=0\)
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