Problem 121 - College Algebra

Question

Find all values of \(x\) satisfying the given conditions. $$\begin{aligned}&y_{1}=2 x^{2}+5 x-4, y_{2}=-x^{2}+15 x-10, \text { and }\\\&y_{1}-y_{2}=0 \end{aligned}$$

Step-by-Step Solution

Answer

The final roots of the equation after calculation would be \(x_1 = 2\) and \(x_2 = 1\)

Step 1: Set Up the Equation

Given that \(y_{1} - y_{2} = 0\), replace \(y_1\) and \(y_2\) with their respective function expressions: \(2x^2 + 5x - 4- (-x^2 + 15x - 10) = 0\)

Step 2: Simplify The Equation

Simplify the equation from step 1, distribute the minus sign to the formation in the parentheses: \(2x^2 + 5x - 4 + x^2 - 15x + 10 = 0\) which results in \(3x^2 -10x + 6 = 0\) when combined like terms.

Step 3: Solve The Equation

This is a quadratic equation in the standard form \(ax^2 + bx + c = 0\), where \(a=3\), \(b=-10\), and \(c=6\). The roots of equation can be found using the quadratic formula \(x= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting \(a=3\), \(b=-10\), and \(c=6\) into the quadratic formula gives you the roots: \(x_{1,2} = \frac{10 \pm \sqrt{(-10)^2 - 4*3*6}}{2*3}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of quadratic equations, which are algebraic expressions of the second degree (in the form of \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants and \(a\) is not zero). To solve such equations using the quadratic formula, one must identify the coefficients \(a\), \(b\), and \(c\) and plug them into the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

As seen in the exercise, the equation is simplified to \(3x^2 -10x + 6 = 0\). Here, \(a=3\), \(b=-10\), \(c=6\). Substituting these values into the quadratic formula allows us to arrive at the solution for \(x\).

It's essential to evaluate the discriminant, \(b^2-4ac\), which determines the nature of the roots. A discriminant greater than zero indicates two distinct real roots, equal to zero indicates a single real root (also called a repeated root), and less than zero indicates that the roots are complex numbers. For the given equation, the discriminant is positive, ensuring two distinct real solutions.

System of Equations

A system of equations is a set of two or more equations that share the same variables, and the solution is the set of values that satisfies all equations simultaneously. In the exercise at hand, we are dealing with a single variable system created by setting two expressions equal to each other. This is done by understanding that the condition \(y_1 - y_2 = 0\) implies both expressions represent the same point on a graph at which they intersect.

To solve the system, we combine the expressions for \(y_1\) and \(y_2\), which are quadratic equations, into a single equation. When they are set equal, we can solve for \(x\) using algebraic methods, such as simplification and the quadratic formula.

Solving systems accurately requires attention to algebraic manipulation, like distributing negative signs correctly and combining like terms, as seen in steps 1 and 2 of the solution.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation symbols. In the context of the problem given, \(2x^2 + 5x - 4\) and \(-x^2 + 15x - 10\) are algebraic expressions representing the values of \(y_1\) and \(y_2\) respectively.

Constructing and simplifying algebraic expressions are crucial skills in solving quadratic equations and systems of equations. This involves the processes of distributing signs across terms (step 2), combining like terms, and organizing the expression into a recognizable form, such as the standard form of a quadratic equation, before applying specific techniques like the quadratic formula for solutions.

Grasping the nature of these expressions and working accurately with them paves the way for solving more complex algebraic problems. Remember, that practice in manipulating these expressions leads to a stronger understanding and quicker recognition of how to approach and solve them.