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Chapter 3: Problem 65

Show that \((23-8 \sqrt{7})^{1 / 2}=4-\sqrt{7}\).

Short Answer

Expert verified
Squaring the expression \((4-\sqrt{7})\) and simplifying it, we obtain \(23-8\sqrt{7}\). This confirms that \((23-8 \sqrt{7})^{1 / 2}=4-\sqrt{7}\).

Step by step solution

01

Square the given expression

To begin, let's square the expression \((4-\sqrt{7})\): \((4-\sqrt{7})^2\)
02

Apply the square of a binomial formula

Now, we will apply the square of a binomial formula, which is given by: \((a-b)^2 = a^2 - 2ab + b^2\) In our case, \(a = 4\) and \(b = \sqrt{7}\), so we have: \((4-\sqrt{7})^2 = 4^2 - 2(4)(\sqrt{7}) + (\sqrt{7})^2\)
03

Simplify the expression

Next, let's simplify the expression: \(16 - 8\sqrt{7} + 7\)
04

Combine like terms

Now, combine the constants: \(16 + 7 = 23\) So, we have: \(23 - 8\sqrt{7}\)
05

Compare the results

As we can see, the square of \((4-\sqrt{7})\) is indeed equal to \((23-8\sqrt{7})\), so it can be concluded that: \((23-8 \sqrt{7})^{1 / 2}=4-\sqrt{7}\)

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