Find x 1+1.5(square root x^2-1 - 1.5x^2/square root of x^2-1

To solve the equation \( 1 + 1.5\left(\sqrt{x^2 - 1} - \frac{1.5x^2}{\sqrt{x^2 - 1}}\right) = 0 \), we can first simplify the expression inside the parentheses.

1. Rewrite the equation:
   \[
   1 + 1.5\left(\sqrt{x^2 - 1} - \frac{1.5x^2}{\sqrt{x^2 - 1}}\right) = 0
   \]

2. Simplify the expression:
   \[
   \sqrt{x^2 - 1} - \frac{1.5x^2}{\sqrt{x^2 - 1}} = \frac{(\sqrt{x^2 - 1})^2 - 1.5x^2}{\sqrt{x^2 - 1}} = \frac{x^2 - 1 - 1.5x^2}{\sqrt{x^2 - 1}} = \frac{-0.5x^2 - 1}{\sqrt{x^2 - 1}}
   \]

3. Substitute back into the equation:
   \[
   1 + 1.5 \cdot \frac{-0.5x^2 - 1}{\sqrt{x^2 - 1}} = 0
   \]

4. Multiply through by \(\sqrt{x^2 - 1}\) to eliminate the fraction:
   \[
   \sqrt{x^2 - 1} + 1.5(-0.5x^2 - 1) = 0
   \]

5. Distribute:
   \[
   \sqrt{x^2 - 1} - 0.75x^2 - 1.5 = 0
   \]

6. Rearranging gives:
   \[
   \sqrt{x^2 - 1} = 0.75x^2 + 1.5
   \]

7. Square both sides:
   \[
   x^2 - 1 = (0.75x^2 + 1.5)^2
   \]

8. Expand the right-hand side:
   \[
   (0.75x^2 + 1.5)(0.75x^2 + 1.5) = 0.5625x^4 + 2.25x^2 + 2.25
   \]

9. Set the equation:
   \[
   x^2 - 1 = 0.5625x^4 + 2.25x^2 + 2.25
   \]

10. Rearranging this gives:
    \[
    0.5625x^4 + 1.25x^2 + 3.25 = 0
    \]

This is a quadratic in terms of \(y = x^2\):
\[
0.5625y^2 + 1.25y + 3.25 = 0
\]

Now you can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find \(y\).

### Calculation of \(y\):
1. \(a = 0.5625\)
2. \(b = 1.25\)
3. \(c = 3.25\)

Calculating the discriminant:
\[
b^2 - 4ac = (1.25)^2 - 4(0.5625)(3.25)
\]

Calculating further gives:
\[
= 1.5625 - 7.35 = -5.7875
\]

Since the discriminant is negative, there are no real solutions for \(x^2\), which means there are no real values for \(x\).

If you need further details or have any questions, feel free to ask! Here are some related questions:

1. How do you interpret complex solutions in equations?
2. Can you provide examples of similar equations with real solutions?
3. What steps can you take when faced with a negative discriminant?
4. How does squaring both sides affect the solutions of an equation?
5. What are the implications of having no real solutions in a mathematical model?

**Tip:** Always check for extraneous solutions after squaring both sides of an equation, as this can introduce false solutions.

Find x 1+1.5(square root x^2-1 - 1.5x^2/square root of x^2-1)

Explore the step-by-step solution to solving a complex equation with radical and quadratic terms, involving discriminant analysis and square root simplification. This guide demonstrates how to handle equations with no real solutions, ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation