Problem 15 - Algebra and Trigonometry

Question

\(15-18\) . Two equations and their graphs are given. Find the inter- section point(s) of the graphs by solving the system. $$ \left\\{\begin{array}{l}{x^{2}+y=8} \\ {x-2 y=-6}\end{array}\right. $$

Step-by-Step Solution

Answer

The intersection points are (2, 4) and (-2.5, 1.75).

Step 1: Identify and Rearrange Equations

The system of equations is given as:\[\begin{align*}x^2 + y &= 8 \x - 2y &= -6\end{align*}\]First, we'll express \(y\) in terms of \(x\) using the second equation because it is linear (easier to rearrange). Rearranging the second equation, we get:\[y = \frac{x + 6}{2}\]

Step 2: Substitute Expression for y in First Equation

Substitute \(y = \frac{x + 6}{2}\) into the first equation, \(x^2 + y = 8\). This gives us:\[x^2 + \frac{x + 6}{2} = 8\]To eliminate the fraction, multiply the entire equation by 2:\[2x^2 + (x + 6) = 16\]

Step 3: Simplify and Solve for x

Now, simplify the equation:\[2x^2 + x + 6 = 16\]Subtract 16 from both sides:\[2x^2 + x + 6 - 16 = 0 \2x^2 + x - 10 = 0\]Solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = 1\), \(c = -10\).

Step 4: Apply the Quadratic Formula

The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substituting the values, we get:\[x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-10)}}{2(2)}\]\[x = \frac{-1 \pm \sqrt{1 + 80}}{4}\]\[x = \frac{-1 \pm \sqrt{81}}{4}\]\[x = \frac{-1 \pm 9}{4}\]Thus, the solutions for \(x\) are \(x = 2\) and \(x = -2.5\).

Step 5: Find Corresponding y Values

For \(x = 2\), substitute back into \(y = \frac{x + 6}{2}\):\[y = \frac{2 + 6}{2} = 4\]For \(x = -2.5\), substitute back:\[y = \frac{-2.5 + 6}{2} = \frac{3.5}{2} = 1.75\]

Step 6: State Intersection Points

The solutions representing the points of intersection are \((2, 4)\) and \((-2.5, 1.75)\).

Key Concepts

Quadratic Equations

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \). This means that it involves an variable 'x' squared, and it is a polynomial equation of degree 2.A characteristic feature of quadratic equations is their graph, which often takes the shape of a U-shaped curve called a parabola.
The critical aspects of quadratic equations include:

• The coefficient 'a', which determines the parabola's direction and width.
• The solutions or roots of the equation, which can be found using various methods like factoring, completing the square, or applying the quadratic formula.
• The nature of the roots, which depends on the discriminant \( b^2 - 4ac \). If the discriminant is positive, the equation has two distinct real roots; if zero, it has one real root; and if negative, the roots are complex.

To solve a quadratic equation in our exercise, we applied the quadratic formula. Remember, it's important to identify 'a', 'b', and 'c' correctly, as these values are crucial for finding the solutions.

Linear Equations

Linear equations are the simplest form of equations, represented as \( ax + by = c \). These equations show a straight-line relationship between the variables 'x' and 'y'.They are classified as polynomials of degree 1.
Key features of linear equations include:

• The slope-intercept form \( y = mx + b \) where 'm' is the slope, indicating how steep the line is, and 'b' is the y-intercept.
• The graph of a linear equation is a line on the Cartesian plane.
• The coefficients of the variables directly impact the slope and positioning of the line.

In the context of systems of equations, linear equations are crucial because they often provide simple expressions for substitution, as observed in the exercise where we expressed 'y' in terms of 'x' for substitution in the quadratic equation.

Intersection Points

Intersection points are the coordinates where two graphs meet or cross each other. In the context of solving a system of equations, finding the intersection points means determining the values of 'x' and 'y' that simultaneously satisfy both equations.
To find these points:

• First, express one of the variables in terms of the other using one equation (preferably the simpler, linear one).
• Substitute this expression into the other equation to create a single-variable equation.
• Solve this equation to find the specific values of 'x'.

Once you have the 'x' values, substitute them back to find the corresponding 'y' values.Intersection points are where the solutions of both equations coincide, revealing crucial insights about the system, such as whether they share one or more common solutions.In our exercise, the curves intersect at points \((2, 4)\) and \((-2.5, 1.75)\), providing two distinct solutions for the system.