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When solving problems involving geometric figures like triangles, often, quadratic equations come into play. A quadratic equation is any equation that can be expressed in the form:\[ax^2 + bx + c = 0\] where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the unknown variable. In this specific triangle problem, we arrived at a quadratic equation while calculating the unknown base of the triangle. - Step 1: We transformed the expression for area with the given height relation into an equation involving \(b\).- Step 2: Rearranging this into a standard quadratic form enabled us to efficiently solve for \(b\) using the quadratic formula.The quadratic formula is a powerful method for finding the roots of quadratic equations:\[x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\] This formula yields two solutions, which may be real or complex numbers, and providing these solutions gives us an insightful result regarding the variable in question. In our exercise, the correct solution based on the context is the positive value for the base as the geometric dimensions cannot be negative.

One of the fundamental concepts in triangle geometry is determining the area. The area of a triangle is calculated using the formula:\[A = \frac{1}{2} \times \text{base} \times \text{height}\] This formula is derived from the fact that the area of a triangle is essentially half the area of a rectangle formed by the same base and height. In the given problem, the area was 60 square units. Therefore, we set up the equation:\[60 = \frac{1}{2} \times b \times h\] Given the relationship between the base and the height, this formula became an equation with one variable, which helped us solve the problem.Understanding this fundamental formula is crucial because it breaks down a triangle into understandable parts, making it easier to analyze. Whether you're dealing with a right triangle, an isosceles triangle, or any other type, this formula is applicable as long as you have a base and a height that intersect perpendicularly.

Algebraic expressions such as \(3b + 2\) play a crucial role in formulating and solving geometry problems. Essentially, these expressions allow us to describe complex relationships in a simplified algebraic form.- **Expression Creation:** In the triangle problem at hand, we wrote the expression for the height \(h\) in terms of the base \(b\) as \(h = 3b + 2\). This expression stemmed from knowing that the height is 2 units more than 3 times the base.- **Substituting in Formulas:** By substituting this expression into our triangle area formula, we formed a new equation in terms of \(b\). This manipulation is pivotal, as solving for \(b\) hinges upon how the expression seamlessly integrates into known formulas.The manipulation of algebraic expressions requires understanding how elements within the problem relate. In geometry, recognizing these relationships allows you to effectively turn descriptive problems into solvable equations. This skill is not only useful in educational settings but also in everyday applications where mathematical relationships and equations provide deep insights into structural and spatial problems.