Problem 19 - Algebra 2 and Trigonometry

Question

In \(19-25,\) express each answer in simplest radical form. Check each answer. The larger of two numbers is 5 more than twice the smaller. The square of the smaller is equal to the larger. Find the numbers.

Step-by-Step Solution

Answer

The smaller number is \(1 + \sqrt{6}\) and the larger number is \(7 + 2\sqrt{6}\).

Step 1: Define the Variables

Let the smaller number be \( x \) and the larger number be \( y \). This helps us to express the relationships given in the problem using algebraic equations.

Step 2: Set Up the Equations

According to the problem, "the larger of two numbers is 5 more than twice the smaller," which gives us the equation \( y = 2x + 5 \). Additionally, "the square of the smaller is equal to the larger" gives us the equation \( x^2 = y \).

Step 3: Substitute Equation

Substitute the expression for \( y \) from the first equation into the second equation. This means we replace \( y \) with \( 2x + 5 \) in the equation \( x^2 = y \), leading to: \( x^2 = 2x + 5 \).

Step 4: Rearrange the Equation

Rearrange the equation by moving all terms to one side so that it becomes a standard quadratic equation: \( x^2 - 2x - 5 = 0 \).

Step 5: Solve the Quadratic Equation

Use the quadratic formula to solve for \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1 \), \( b = -2 \), and \( c = -5 \). Calculate the discriminant: \( (-2)^2 - 4 \cdot 1 \cdot (-5) = 4 + 20 = 24 \). Thus, \[ x = \frac{2 \pm \sqrt{24}}{2} \].

Step 6: Simplify the Solution

Simplify \( \sqrt{24} \) to \( 2\sqrt{6} \), so the solutions become \[ x = 1 \pm \sqrt{6} \]. Since \( x \) is a number, we choose only the positive values: \( x = 1 + \sqrt{6} \).

Step 7: Find the Larger Number

Substitute \( x = 1 + \sqrt{6} \) back into the equation \( y = 2x + 5 \) to find \( y \). Then, \[ y = 2(1 + \sqrt{6}) + 5 = 2 + 2\sqrt{6} + 5 = 7 + 2\sqrt{6} \].

Step 8: Verification

Verify the solution by checking that \( x^2 = y \). Calculate \( (1 + \sqrt{6})^2 = 1 + 2\sqrt{6} + 6 = 7 + 2\sqrt{6} \), which matches \( y \). Therefore, the solution satisfies both original equations.

Key Concepts

Radical Expressions

Radical expressions involve square roots, cube roots, or any other root expressions. These forms allow us to express values in terms of roots, providing a clearer representation of numbers that are not perfect squares, cubes, etc.

When solving quadratic equations like the one in our problem

• We encountered roots while simplifying, particularly when determining values such as \( \sqrt{24} \).
• Simplifying these radicals further, \( \sqrt{24} \) can be broken down into \( 2\sqrt{6} \), since 24 can be factored into \( 4 \times 6 \).

This simplification helps in keeping solutions neat and understandable, ensuring all parts of the equation remain in their most elementary radical form.

Completing the Square

Completing the square is a method used to solve quadratic equations and gives insights into their structure. Although this particular exercise primarily employed the quadratic formula, completing the square is an equally powerful method.It involves

• Rewriting the equation in the form \((x - p)^2 = q\).
• This allows you to derive the roots directly.

For our quadratic equation

• \(x^2 - 2x - 5 = 0\), instead of initially using the quadratic formula, you could bring all constant terms to one side and transform it.
• By adding and subtracting the square of half the coefficient of \(x\), the equation reforms into a perfect square trinomial.

Although this wasn't applied in the given solution, knowing this method could be beneficial when the problem demands a more versatile approach.

Algebraic Manipulation

Algebraic manipulation is key in transforming and solving equations. It starts with identifying relationships given in word problems and translating them into mathematical expressions.

• For example, creating those initial equations: \( y = 2x + 5 \) and \( x^2 = y \).
• This requires rearranging terms and solving systems of equations.

When we substituted the expression for \(y\) in terms of \(x\), we used algebraic manipulation to consolidate variables, deriving \( x^2 = 2x + 5 \).

• This led us to a solvable quadratic equation for \(x\).
• Ensuring each step in algebraic manipulation is precise prevents errors.

This is especially crucial when expressing answers in simpler forms like radicals or verifying the correctness of an answer.

Verification of Solutions

Verification is a crucial step in ensuring that the solutions obtained are correct according to the original problem.

• This involves substituting the found solutions back into the original equations.
• For our problem, after finding \(x = 1 + \sqrt{6}\), and corresponding \(y = 7 + 2\sqrt{6}\), substitution showed that both original conditions were satisfied.

By calculating \((1 + \sqrt{6})^2\), we see that it equals \(7 + 2\sqrt{6}\), matching precisely with our expression for \(y\).

• This confirmed the accuracy of our solution.

Verification not only proves correctness but also builds confidence in solving more complex equations in the future.