The equation you've provided is:

\[
\frac{x^2}{5} + \frac{1}{2} = \frac{x}{3} + \frac{5}{8}
\]

Let's solve it step by step.

### Step 1: Eliminate the fractions

We can get rid of the denominators by multiplying the entire equation by the least common denominator (LCD). The LCD of 5, 2, 3, and 8 is 120.

So, multiply both sides of the equation by 120:

\[
120 \cdot \left(\frac{x^2}{5}\right) + 120 \cdot \left(\frac{1}{2}\right) = 120 \cdot \left(\frac{x}{3}\right) + 120 \cdot \left(\frac{5}{8}\right)
\]

This simplifies to:

\[
24x^2 + 60 = 40x + 75
\]

### Step 2: Rearrange the equation

Now, move all the terms to one side to set the equation to zero:

\[
24x^2 - 40x + 60 - 75 = 0
\]

Simplifying:

\[
24x^2 - 40x - 15 = 0
\]

### Step 3: Solve the quadratic equation

This is a quadratic equation in the form:

\[
24x^2 - 40x - 15 = 0
\]

We can solve this using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \(a = 24\), \(b = -40\), and \(c = -15\).

Let's calculate the discriminant (\(b^2 - 4ac\)):

\[
\Delta = (-40)^2 - 4(24)(-15) = 1600 + 1440 = 3040
\]

Now, applying the quadratic formula:

\[
x = \frac{40 \pm \sqrt{3040}}{48}
\]

Let's simplify further by calculating the square root of 3040.

The square root of 3040 is approximately \( 55.14 \).

Now, applying the quadratic formula:

\[
x = \frac{40 \pm 55.14}{48}
\]

This gives two possible solutions:

1. \( x_1 = \frac{40 + 55.14}{48} = \frac{95.14}{48} \approx 1.98 \)
2. \( x_2 = \frac{40 - 55.14}{48} = \frac{-15.14}{48} \approx -0.32 \)

Thus, the two possible solutions for \(x\) are:

\[
x_1 \approx 1.98 \quad \text{and} \quad x_2 \approx -0.32
\]

Would you like more details on any step, or have any questions?

Here are 5 related questions:
1. What is the quadratic formula, and why does it work for all quadratic equations?
2. How do we find the least common denominator (LCD) for fractions?
3. Can quadratic equations always be solved by factoring?
4. How can we interpret the discriminant in a quadratic equation?
5. What are some real-world examples of quadratic equations?

**Tip**: In quadratic equations, if the discriminant is positive, there are two real solutions; if it's zero, there's one real solution; and if negative, the solutions are complex.

This detailed step-by-step guide explains how to solve a quadratic equation involving fractions: X²/5 + 1/2 = x/3 + 5/8. Learn key algebraic concepts, the quadratic equation formula, and the method for eliminating fractions. Ideal for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation