Question

1. write a solution to the equation $x^2 = 5$ two ways:

a. using exponents
b. using radicals

1. write $sqrt3{78}$ using an exponent instead of a radical.

Explanation:

Problem 1

Part a: Using exponents

Step1: Recall exponent rule for roots

To solve \(x^{2}=5\) for \(x\), we use the property that if \(x^{n}=a\), then \(x = a^{\frac{1}{n}}\) (for real - valued solutions when \(n\) is even and \(a\geq0\)). Here \(n = 2\) and \(a = 5\).
So \(x=5^{\frac{1}{2}}\) (we can also consider the negative solution \(x=- 5^{\frac{1}{2}}\) since \((-5^{\frac{1}{2}})^{2}=5\) as well)

Part b: Using radicals

Step1: Recall radical notation

The square root of a number \(a\) is denoted as \(\sqrt{a}\), and for the equation \(x^{2}=5\), by the definition of square root, if \(x^{2}=a\) (\(a\geq0\)), then \(x = \pm\sqrt{a}\).
So \(x=\pm\sqrt{5}\)

Problem 2

Step1: Recall the relationship between radicals and exponents

The \(n\) - th root of a number \(a\), \(\sqrt[n]{a}\), can be written as \(a^{\frac{1}{n}}\) in exponential form. Here, we have the cube root of \(78\), so \(n = 3\) and \(a = 78\).
So \(\sqrt[3]{78}=78^{\frac{1}{3}}\)

Final Answers

1. a. \(x = 5^{\frac{1}{2}}\) (or \(x=-5^{\frac{1}{2}}\))
2. b. \(x=\pm\sqrt{5}\)
3. \(78^{\frac{1}{3}}\)

Answer:

Step1: Recall the relationship between radicals and exponents

The \(n\) - th root of a number \(a\), \(\sqrt[n]{a}\), can be written as \(a^{\frac{1}{n}}\) in exponential form. Here, we have the cube root of \(78\), so \(n = 3\) and \(a = 78\).
So \(\sqrt[3]{78}=78^{\frac{1}{3}}\)

Final Answers

1. a. \(x = 5^{\frac{1}{2}}\) (or \(x=-5^{\frac{1}{2}}\))
2. b. \(x=\pm\sqrt{5}\)
3. \(78^{\frac{1}{3}}\)