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Chapter 3: Problem 58

Find an integer \(m\) such that $$ \left((5-2 \sqrt{3})^{2}-m\right)^{2} $$ is an integer.

Short Answer

Expert verified
The integer value of \(m\) that makes the given expression an integer is \(m = 37 - 40\sqrt{3}\).

Step by step solution

01

Expand inner square

First, expand the inner square by squaring the binomial \((5-2\sqrt{3})^2\): \((5-2\sqrt{3})^2 = (5-2\sqrt{3})(5-2\sqrt{3})\) Remember to apply the FOIL method (first, outer, inner, last).
02

Apply the FOIL method

Multiply the terms from the binomials together: \((5)(5) - (5)(2\sqrt{3}) - (2\sqrt{3})(5) + (2\sqrt{3})(2\sqrt{3})\) Now, simplify by combining like terms.
03

Simplify the expression

Combine and simplify the expression: \(25 - 10\sqrt{3} - 10\sqrt{3} + 12 = 37 - 20\sqrt{3}\) Now that we have the simplified expression, we can plug it back into the original expression.
04

Plug the simplified expression back in

Now insert the simplified inner square expression back into the original expression: \(\left((37-20\sqrt{3}) - m\right)^2\) Now, we need to find an integer value for \(m\) that will make this expression an integer.
05

Determine the integer value for m

In order to have an integer result, the content inside the square must be a perfect square. Let's check when the square root of the content will not introduce another irrational term. By doing so, we check how our \(m\) will nullify the irrational term: \(\sqrt{(37-20\sqrt{3})-m} = a\) Solving for \(m\), we get: \(m = 37 - 20\sqrt{3} - a^2\) Now let's focus on the irrational term. We want this part to be nullify: \(-20\sqrt{3} - a^2 = 0\) Solving the equation, we find the value of \(a\): \(a^2 = 20\sqrt{3}\) Now that we have the value of \(a^2\), we can find the value of \(m\).
06

Calculate the value of m

Substitute the value of \(a^2\) into the equation for \(m\): \(m = 37 - 20\sqrt{3} - 20\sqrt{3}\) \(m = 37 - 40\sqrt{3}\) Therefore, the integer value of \(m\) that makes the given expression an integer is \(m = 37 - 40\sqrt{3}\).

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