Question

the values or \((5+2 \sqrt{6})^{x^{2}-3}+(5-2 \sqrt{6})^{x^{2}-3}=10\) \((a) \pm 2 \pm \frac{\sqrt{3}}{2}\) (b) \(\pm \sqrt{2}, \pm 4\) (d) \(2 \sqrt{2}, \sqrt{3}\)

Short answer

Short answer: None of the given options are the solution to the given equation \((5+2 \sqrt{6})^{x^{2}-3}+(5-2 \sqrt{6})^{x^{2}-3}=10\), as the sum of their respective \(A^{2y}\) and \(B^{2y}\) values does not equal 98.

Step-by-step solution

Rewrite the expression

We can rewrite the given equation as \(A^{x^2-3}+B^{x^2-3}=10\), with \(A=5+2\sqrt{6}\) and \(B=5-2\sqrt{6}\).

Work with the conjugate pairs

Since \(A\) and \(B\) are conjugate pairs, we can write the following: \(AB=(5+2\sqrt{6})(5-2\sqrt{6})=1\).

Introduce a new variable

Let \(y=x^2-3\), then the given equation becomes \(A^{y}+B^{y}=10\). Now, we can write the equation as \(B^{y}+A^{y}=10\), and multiply both equations. \(A^{y}+B^{y}=10\) \(B^{y}+A^{y}=10\)

Multiply the equations

Multiply both equations to obtain \((A^{y}+B^{y})(B^{y}+A^{y})=100\). Substitute \(AB=1\) from Step 2, we have: \((A^{y}B^{y}+A^{2y}+B^{2y}+A^{y}B^{y})=100\) \((1+A^{2y}+B^{2y}+1)=100\)

Simplify and solve for y

Simplify the above equation to get: \(A^{2y}+B^{2y}=98\) Now we can compare the equation with the options given and find which amongst them can fit as the value of \(y\).

Check the options

Option (a): Using \(\pm 2 \pm \frac{\sqrt{3}}{2}\) as the values for y will not produce a sum of 98. Option (b): If y = \(\pm \sqrt{2}\), the sum will be \((5+2\sqrt{6})^{2}+(5-2\sqrt{6})^{2} \neq 98\). Option (d): If y = \(2\sqrt{2}\) or \(\sqrt{3}\), the sum will be \((5+2\sqrt{6})^{2\sqrt{2}}+(5-2\sqrt{6})^{2\sqrt{2}} \neq 98\) We can see that none of the options are giving a sum equal to 98. Therefore, none of the given options are the solution to the given equation.

Key concepts

Conjugate Pairs

When dealing with complex algebraic expressions, conjugate pairs are a powerful tool for simplification. A conjugate pair consists of two binomials that are identical except for the sign between their terms. For instance, if we have a binomial like \(a + b\), its conjugate is \(a - b\). One of the most remarkable properties of conjugate pairs is that when they are multiplied together, the product is always the difference of squares.

This method exploits the fact that the cross terms in the multiplication of conjugate pairs cancel each other out, simplifying the multiplication to \( a^2 - b^2 \). In our example, working with conjugates simplified a term to 1, highlighting the practical power of this technique. Recognizing and utilizing conjugate pairs in equations is an essential strategy to streamline complex algebraic problems and solve them more efficiently.

Variable Substitution

Working through complicated equations can sometimes feel like navigating a maze, but variable substitution can be the secret passage that simplifies the journey. This algebraic technique involves replacing a more complex expression with a single variable to make the equation easier to manage. It is akin to temporarily renaming a part of the equation.

After completing operations with the simpler form, one can then 'unsubstitute' and revert to the original variables to solve the problem. Our example uses variable substitution effectively by setting \(y = x^2 - 3\), which dramatically streamlines the equation. As we then focus on solving for \(y\), the process becomes much more manageable. This down-shifts the complexity from a more perplexing equation to an easier, more approachable one, showcasing how variable substitution is a vital tactic in the algebraic toolkit.

Exponents

Powerful and deceptive in their simplicity, exponents are the succinct shorthand that allows us to express repeated multiplication compactly. The exponent of a number tells us how many times to multiply the number by itself. For instance, \(x^3\) means \(x \times x \times x\). But with great power comes great complexity; the rules governing the operations involving exponents require careful attention.

When solving equations, we often encounter situations where we must apply the laws of exponents — such as multiplying and dividing powers, raising a power to a power, or dealing with zero or negative exponents. In our example, we manipulated exponentiated variables and ultimately sought a value of \(y\) that would satisfy a specific numerical result when applied as an exponent to given bases. Deep understanding of how exponents work allows us to untangle such intricate expressions and reveal the solution hidden within the exponential growth or decay.