Problem 39 - Algebra 2 and Trigonometry

Question

The lengths of the legs of an isosceles right triangle are \(x\) and \(\sqrt{6 x+16} .\) What are the lengths of the legs of the triangle?

Step-by-Step Solution

Answer

The lengths of the legs of the triangle are both 8.

Step 1: Understanding the Properties of Isosceles Right Triangle

An isosceles right triangle has two sides of equal length and one angle that is 90 degrees. If we denote the legs of the triangle as "\(x\)" and "\(\sqrt{6x + 16}\)," they must be equal because they're the legs of an isosceles triangle.

Step 2: Set the Legs Equal

Equate the expressions for the legs of the triangle according to the property of the isosceles triangle. So, we have: \[ x = \sqrt{6x + 16} \]

Step 3: Square Both Sides to Eliminate the Square Root

Square both sides of the equation to eliminate the square root: \[ x^2 = 6x + 16 \]

Step 4: Rearrange the Equation into Standard Quadratic Form

Reorganize the equation into the standard form of a quadratic equation: \[ x^2 - 6x - 16 = 0 \]

Step 5: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = -16\), to find the values of \(x\): Calculate the discriminant: \[ b^2 - 4ac = (-6)^2 - 4 \times 1 \times (-16) = 36 + 64 = 100 \] Apply the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{100}}{2 \times 1} = \frac{6 \pm 10}{2} \]This gives two potential solutions for \(x\): \[ x = \frac{6 + 10}{2} = 8 \] \[ x = \frac{6 - 10}{2} = -2 \]

Step 6: Determine the Valid Value for \(x\)

Since \(x\) represents a length, it must be positive, so we discard \(-2\) as a solution. Thus, \(x = 8\) is the valid length.

Step 7: Verify Solution

Plug \(x = 8\) back into the equation for the other leg: \[ \sqrt{6x + 16} = \sqrt{6(8) + 16} = \sqrt{48 + 16} = \sqrt{64} = 8 \]Since both legs are equal, both legs of the isosceles right triangle are 8.

Key Concepts

Quadratic Equation

Quadratic equations are a cornerstone of algebra, often encountered in the context of various geometric problems. These equations take the standard form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable we aim to determine.
One of the most important tools for solving quadratic equations is the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• The formula offers two solutions due to the "\( \pm \)" symbol, which represents both the addition and subtraction of the square root term.
• If the "discriminant" \( b^2 - 4ac \) is positive, there are two real and distinct solutions.
• If it is zero, the equation has exactly one real solution.
• A negative discriminant means there are no real solutions.

In our exercise, the quadratic equation \( x^2 - 6x - 16 = 0 \) is solved using this formula. With \( a = 1 \), \( b = -6 \), and \( c = -16 \), we found two possible values for \( x \), of which only the positive \( x = 8 \) is physically meaningful as a length.

Square Root

The square root operation helps to "undo" the squaring of a number, making it a frequent tool for solving geometric problems.
When dealing with a square root, like \( \sqrt{6x + 16} \), we encounter expressions where the value is raised to a power of one-half. The square root asks what number, when squared, will yield the value inside the root.
For solving equations with a square root, a common step is to eliminate it by squaring both sides. This method allows for the application of algebraic manipulations needed to solve for the variable.

• Always be cautious: squaring both sides can introduce extraneous solutions
• These solutions might be non-physical in given contexts, like negative lengths in geometry
• Verification is key: Always substitute back to check potential solutions

Thus, it is important to verify solutions, as was done in the exercise, ensuring the solution is consistent with the geometric properties of the problem.

Triangular Geometry

Triangular geometry plays a vital role in understanding various properties and applications within mathematics. In particular, isosceles right triangles offer a unique set of properties that make them interesting.
An isosceles right triangle is defined as having two sides of equal length and one right angle, measuring 90 degrees. These triangles are special because they combine simplicity with specific ratios of sides and angles.

• Such triangles always have equal lengths for the two legs opposite the congruent angles
• The hypotenuse is typically \( x\sqrt{2} \) if the legs are \( x \)
• For any isosceles triangle, checking the equality of lengths confirms the triangle's identity

In our exercise, the triangle's two legs are expressed in terms of \( x \) and \( \sqrt{6x + 16} \). Equating these confirmed their equality, establishing the triangular relationships and leading to the solution through algebraic manipulation.