Question

The roots of the equation \((a+\sqrt{b})^{x^{2}-15}+(a-\sqrt{b})^{2^{2}-15}=2 a\), where \(a^{2}-b=1\) are (a) \(\pm 6, \pm \sqrt{20}\) (b) \(\pm 3, \pm \sqrt{5}\) (c) \(\pm 4, \pm \sqrt{14}\) (d) \(\pm 2, \pm 3\)

Short answer

Answer: The roots of the equation are (c) \(\pm 4\).

Step-by-step solution

Write the given equation in a simplified form

We are given the equation \((a+\sqrt{b})^{x^{2}-15}+(a-\sqrt{b})^{2^{2}-15}=2a\). It can be simplified to \((a+\sqrt{b})^{x^{2}-15}+(a-\sqrt{b})^{4-15}=2a\), and further simplified to \((a+\sqrt{b})^{x^{2}-15}+(a-\sqrt{b})^{-11}=2a\).

Use the given condition \(a^{2}-b=1\)

We are given \(a^{2}-b=1\). Let's solve for \(\sqrt{b}\): \(\sqrt{b}=\sqrt{a^2-1}\). Now substitute this into the equation \((a+\sqrt{b})^{x^{2}-15}+(a-\sqrt{b})^{-11}=2a\): \((a+\sqrt{a^2-1})^{x^{2}-15}+(a-\sqrt{a^2-1})^{-11}=2a\).

Make a substitution to simplify the equation

Let's make a substitution: let \(\cosh{u}=a\) and \(\sinh{u}= \sqrt{a^2-1}\). This means that \(a-\sinh{u}=\cosh{u}-\sinh{u}\) and \(a+\sinh{u}=\cosh{u}+\sinh{u}\). The equation becomes: \((\cosh{u} + \sinh{u})^{x^2-15}+(\cosh{u}-\sinh{u})^{-11}=2\cosh{u}\).

Apply the hyperbolic trigonometric identity

We know that \(\cosh{u}+\sinh{u}=e^u\) and \(\cosh{u}-\sinh{u}=e^{-u}\). Substitute this into the equation: \((e^u)^{x^2-15}+(e^{-u})^{-11}=2\cosh{u}\).

Simplify the equation

After substituting the hyperbolic trigonometric identity, the equation becomes: \(e^{(x^2-15)u}+e^{11u}=2\cosh{u}\).

Find the value of \(x^2\)

We know that \(2\cosh{u}=e^u+e^{-u}\). Now, let's compare the given equation and the definition of hyperbolic cosine: \(e^{(x^2-15)u}+e^{11u}=e^u+e^{-u}\). To make both sides equal, we get \(x^2-15=1\) and \(11u=-u\). Thus, \(x^2=16\).

Find the roots of the equation

From \(x^2=16\), we can find the roots: \(x=\pm \sqrt{16}\) \(x=\pm 4\). Hence, the roots of the equation are (c) \(\pm 4, \pm \sqrt{14}\).

Key concepts

Hyperbolic Trigonometric Functions

Hyperbolic trigonometric functions are analogs of the ordinary trigonometric functions like sine and cosine, but for the hyperbola instead of the circle. They are defined using exponential functions and can be very useful in solving various mathematical problems, especially those involving exponential equations.

Some basic hyperbolic functions include:

• Hyperbolic sine: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• Hyperbolic cosine: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions satisfy similar identities to their trigonometric counterparts, such as:

• \( \cosh^2(x) - \sinh^2(x) = 1 \)

This identity is essential in many mathematical applications because it relates \( \cosh \) and \( \sinh \) in a way similar to the Pythagorean theorem used with cosine and sine.

In the context of solving the provided problem, we use hyperbolic identities to simplify the given equation and make substitutions that allow other mathematical techniques, such as solving polynomial equations, to be applied effectively.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra that involves finding the values of \(x\) that make a quadratic equation true. A quadratic equation is any equation that can be written in the standard form:
\[ ax^2 + bx + c = 0 \]
where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable.

To solve the quadratic equation, the quadratic formula is commonly used:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula provides the solutions or 'roots' of the quadratic equation. The term under the square root, \(b^2 - 4ac\), is known as the discriminant. It can tell us how many real solutions the equation has based on its value. If:

• \(b^2 - 4ac > 0\), there are two distinct real roots.
• \(b^2 - 4ac = 0\), there is one real root.
• \(b^2 - 4ac < 0\), there are no real roots.

In this particular exercise, resolving the quadratic equation \(x^2 = 16\) gives the roots of \(x = \pm 4\), as shown in the step-by-step solution.

Substitution Method

The substitution method is a technique used to simplify a mathematical problem by replacing a complex expression with a simpler one. This method is particularly beneficial in complex equations or systems where directly solving would be cumbersome.

In the given exercise, substitution is used to transform the equation into a more manageable form by introducing new variables that replace complex expressions. Here, we set \( \cosh{u} = a \) and \( \sinh{u} = \sqrt{a^2-1} \). This substitution leverages the identities:

• \( \cosh{u} + \sinh{u} = e^u \)
• \( \cosh{u} - \sinh{u} = e^{-u} \)

Substituting back into the equation helps in aligning terms and applying known identities, eventually leading to a simpler equation that relates directly to familiar forms of mathematical expressions.

Rewriting complex expressions using substitution is instrumental in breaking down the problem into smaller, more solvable parts, ultimately resulting in finding solutions that fit the original problem statement.