Question

If \(x=7+4 \sqrt{3}\) and \(x y=1\), then the value of \(\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)\) is: (b) 57 (a) 194 (d) \(\frac{7+4 \sqrt{3}}{7-4 \sqrt{3}}\) (c) \(85 \sqrt{3}\)

Short answer

Answer: \(\frac{194}{3881}\)

Step-by-step solution

Find the value of \(y\)

Given that \(xy = 1\) and \(x = 7 + 4\sqrt{3}\), we can find the value of \(y\) by dividing both sides of the equation by \(x\): \(y = \frac{1}{x} = \frac{1}{7 + 4\sqrt{3}}\) To rationalize the denominator, we can multiply both numerator and denominator by the conjugate of the denominator, which is \((7 - 4\sqrt{3})\): \(y = \frac{1(7 - 4\sqrt{3})}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})}\) Now calculate the denominator using the difference of squares: \(y = \frac{7 - 4\sqrt{3}}{(7)^2 - (4\sqrt{3})^2} = \frac{7 - 4\sqrt{3}}{49 - 48} = 7 - 4\sqrt{3}\)

Substitute values of \(x\) and \(y\) in the expression

Now we know the values of \(x\) and \(y\), so we can substitute them into the expression: \(\frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{(7 + 4\sqrt{3})^2} + \frac{1}{(7 - 4\sqrt{3})^2}\)

Calculate the individual terms

Calculate the squares of \(x\) and \(y\): \((7 + 4\sqrt{3})^2 = (7)^2 + 2(7)(4\sqrt{3}) + (4\sqrt{3})^2 = 49 + 56\sqrt{3} + 48 = 97 + 56\sqrt{3}\) \((7 - 4\sqrt{3})^2 = (7)^2 - 2(7)(4\sqrt{3}) + (4\sqrt{3})^2 = 49 - 56\sqrt{3} + 48 = 97 - 56\sqrt{3}\)

Compute the final expression

Now substitute the squares back into the expression and simplify: \(\frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{97 + 56\sqrt{3}} + \frac{1}{97 - 56\sqrt{3}}\) To rationalize the denominators, we can multiply each term by the conjugate of the other term's denominator: \(\frac{1}{97 + 56\sqrt{3}}(1) \cdot \frac{97 - 56\sqrt{3}}{97 - 56\sqrt{3}} + \frac{1}{97 - 56\sqrt{3}}(1) \cdot \frac{97 + 56\sqrt{3}}{97 + 56\sqrt{3}}\) This simplifies to: \(\frac{97 - 56\sqrt{3}}{3881} + \frac{97 + 56\sqrt{3}}{3881} = \frac{194}{3881}\) The final expression is equivalent to option (a) 194.

Key concepts

Algebraic Identities

Algebraic identities are equations that are true for all values of the variables involved. They serve as shortcuts to simplify algebraic expressions and to perform computations quickly and accurately. One of the fundamental identities in algebra is the difference of squares identity, which states that: \[ a^2 - b^2 = (a + b)(a - b) \].
This identity is particularly useful for finding the product of a sum and a difference of the same two terms without actually having to carry out the multiplication. It is also instrumental when we need to rationalize the denominator or simplify algebraic fractions, as seen in the textbook problem involving \[ 7 + 4\sqrt{3} \].
When working with algebraic identities, it's important to recognize patterns that match the identity structure. This recognition allows us to apply the identity directly rather than expand and simplify, which can be much more time-consuming.

Conjugate of a Binomial

The conjugate of a binomial refers to a pair of binomial expressions that only differ in the sign between their terms. If we have a binomial \[ a + b \],
its conjugate is \[ a - b \].
For example, the conjugate of \[ 7 + 4\sqrt{3} \] is \[ 7 - 4\sqrt{3} \].

Using Conjugates to Rationalize the Denominator

When we encounter a radical expression in the denominator, like \[ 4\sqrt{3} \], we can multiply the fraction by a form of one that consists of the conjugate of the denominator over itself. This method, known as rationalizing the denominator, eliminates the radical and typically leads to a simpler expression. Applying this technique was crucial in the given problem to find \[ y \] and further evaluate the expression \[ \frac{1}{x^2} + \frac{1}{y^2} \]. This process also avoids leaving irrational numbers in the denominator, which is standard practice in mathematics to provide clearer and more easily interpretable results.

Difference of Squares

The difference of squares formula is a specific case of the algebraic identities that facilitates the factorization of expressions where one perfect square is subtracted from another. This identity is expressed as: \[ a^2 - b^2 = (a + b)(a - b), \]
which simply breaks down the difference between two squares into the product of their sum and their difference.

Application in Rationalizing the Denominator

The difference of squares plays a pivotal role in rationalizing denominators when we have binomials involving square roots. By multiplying by the conjugate, we're effectively using the difference of squares to simplify the expression. In the exercise, when dealing with \[ y = \frac{1}{7 + 4\sqrt{3}}, \]
we used the conjugate \[ 7 - 4\sqrt{3} \] to transform the denominator into a recognizable difference of squares form, which is then easily simplified. This step is not just a mathematical maneuver; it's a deliberate move to achieve a simpler form for more straightforward calculation and interpretation.