Problem 27 - Algebra 2 and Trigonometry

Question

In \(18-35,\) find each common solution algebraically. Express irrational roots in simplest radical form. $$ \begin{array}{l}{y=x^{2}-2 x+1} \\ {y=-\frac{9 x-35}{2}}\end{array} $$

Step-by-Step Solution

Answer

Common solutions are \((3, 4)\) and \((-5.5, 31.25)\).

Step 1: Set the Equations Equal

Since the two expressions both equal \(y\), set the right-hand sides of the equations equal to each other: \[x^2 - 2x + 1 = -\frac{9x - 35}{2}\].

Step 2: Clear the Fraction

To eliminate the fraction, multiply every term by 2 to get: \[2(x^2 - 2x + 1) = -(9x - 35)\] This simplifies to \[2x^2 - 4x + 2 = -9x + 35\].

Step 3: Move All Terms to One Side

Rearrange the equation to move all terms to one side: \[2x^2 - 4x + 2 + 9x - 35 = 0\] This simplifies to \[2x^2 + 5x - 33 = 0\].

Step 4: Use the Quadratic Formula

Since factoring this quadratic is difficult, use the quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] to solve for \(x\), where \(a = 2\), \(b = 5\), \(c = -33\).

Step 5: Find the Discriminant

Calculate the discriminant, \(b^2 - 4ac\): \[b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-33) = 25 + 264 = 289\].

Step 6: Solve for x

Substitute back into the quadratic formula: \[x = \frac{-5 \pm \sqrt{289}}{4}\]. Since \(\sqrt{289} = 17\), it simplifies to \[x = \frac{-5 \pm 17}{4}\]. This results in two solutions: \[x = \frac{12}{4} = 3\] and \[x = \frac{-22}{4} = -5.5\].

Step 7: Find Corresponding y Values

Substitute each \(x\) value back into either equation to find the corresponding \(y\) values. For \(x = 3\): \[y = 3^2 - 2 \cdot 3 + 1 = 4\]. For \(x = -5.5\): \[y = (-5.5)^2 - 2 \cdot (-5.5) + 1 = 19.25 + 11 + 1 = 31.25\].

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool in algebra for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This powerful formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \),

where \( a \), \( b \), and \( c \) are coefficients of the equation. By substituting the values for \( a \), \( b \), and \( c \) from your quadratic equation, you can find the \( x \) values that satisfy the equation.
In the exercise solution, the quadratic formula was used to solve \( 2x^2 + 5x - 33 = 0 \). By identifying \( a = 2 \), \( b = 5 \), and \( c = -33 \), the solutions for \( x \) could be precisely calculated. The beauty of this formula is its ability to cater to all quadratic equations, regardless of the coefficients.

Discriminant

The discriminant is the part of the quadratic formula under the square root sign, represented as \( b^2 - 4ac \). This value is crucial in determining the nature of the roots of the quadratic equation. Here's how:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root (also known as a repeated root).
• If \( b^2 - 4ac < 0 \), there are no real roots, only complex ones.

For the given problem, calculating the discriminant involved \( b^2 - 4ac = 25 + 264 = 289 \), which is positive. This indicates the equation has two real and distinct roots. Understanding the discriminant allows you to predict the number and type of solutions before fully solving the equation.

Solving Equations

Solving equations involves finding the values of variables that satisfy a given mathematical statement. In algebra, it is common to solve for the variable \( x \) in expressions. Several methods can be used, such as factoring, completing the square, or using the quadratic formula.
When the quadratic equation from the exercise, \( 2x^2 + 5x - 33 = 0 \), was set, it was initially hard to factor. Hence, the quadratic formula was applied for an efficient solution. The goal is to isolate \( x \) and find its specific values that make the equation equal zero. By doing so, we solve the equation and understand the relationship between the variables involved.

Radical Expressions

Radical expressions involve mathematical expressions that use roots, such as square roots (\( \sqrt{} \)). The presence of a radical often complicates an equation, which is why these expressions usually need to be simplified.
During the problem-solving, the radical expression \( \sqrt{289} \) appeared in the quadratic formula. Simplifying it to \( 17 \) was easy since 289 is a perfect square. Simplifying radical expressions aids in further simplifying equations and deriving final answers. It often involves breaking down terms under the square root or calculating precise roots when possible. Handling radical expressions is crucial for problem solving, especially in expressions where obtaining the most simplified form is necessary for clear solutions.