Problem 8 - Algebra 1

Question

Solve the equation. Remember to check for extraneous solutions. $$\frac{x}{6}+\frac{15}{x}=\frac{5}{6}$$

Step-by-Step Solution

Answer

The given equation does not have any real solutions.

Step 1: Clear the fractions

Multiply through by \(6x\) to eliminate the fractions. This yields \(x^2+90=5x\).

Step 2: Rearrange into quadratic form

Rearrange the above equation to get it into a standard quadratic form \(ax^2+bx+c=0\). So, \(x^2-5x+90=0\).

Step 3: Solve the quadratic equation

This equation can be solved either by factoring, completing the square, or using the quadratic formula. Here, the quadratic formula \((-b \pm \sqrt{b^2-4ac})/2a\) gives \(x = (5 \pm \sqrt{5^2-4*1*90})/ (2*1) = (5 \pm \sqrt{25-360})/2\). Since the root of the radicand \(25-360\) is negative and the square root of a negative number does not produce a real number, there are no real solutions.

Step 4: Check for extraneous solutions

Since there are no potential solutions from Step 3, there is no need to check for extraneous solutions.

Key Concepts

Quadratic Equations

Quadratic equations are essential in algebra and appear in various real-world situations, such as calculating areas and determining projectile paths. A quadratic equation is typically in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The presence of \( x^2 \) indicates it's a quadratic equation, as it involves the variable raised to the second power.
Key properties of quadratic equations include their potential to have:

• Two real roots
• One real root
• Complex roots, if the discriminant (the expression \( b^2 - 4ac \)) is less than zero

Quadratics can be solved using various methods: factoring, completing the square, and the quadratic formula. In the context of our exercise, the quadratic equation derived, \( x^2 - 5x + 90 = 0 \), was found to have no real solutions due to a negative discriminant, implying complex roots.

Extraneous Solutions

Extraneous solutions are outcomes that appear in the process of solving an equation but don't actually satisfy the original equation. They often arise in problems involving equations that have been manipulated, like when clearing fractions by multiplying through both sides.
This manipulation can introduce solutions that don't work in the initial equation. In our exercise, though, once we determined there were no real solutions because the quadratic's discriminant was negative, there was no need to check for extraneous solutions.
It's always important to substitute potential solutions back into the original equation to ensure they work, especially in situations where you may have squared both sides or crossed out terms.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying equations to solve them. In our exercise, it was crucial to clear the fractions initially by multiplying the entire equation by \( 6x \). This step simplified the problem into a construct that allowed easier handling as a quadratic equation.
Successfully manipulating an equation involves operations like:

• Adding or subtracting terms
• Multiplying or dividing both sides by the same number
• Factoring expressions
• Using identities and properties of numbers

Finding solutions becomes much more tractable when manipulated into a familiar format. This manipulation is fundamental in providing clarity, especially when faced with multiple algebraic structures mixed together, such as fractions and polynomials.