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What are the possible numbers of solutions for equations in the form [tex]\( ax = \frac{k}{x} \)[/tex]?

How about equations like [tex]\( ax + b = \frac{1}{x} \)[/tex]?

How are these equations related to quadratic equations?


Sagot :

Certainly! Let's analyze the given types of equations step-by-step and determine the possible number of solutions for each.

### 1. Equation of the Form [tex]\( a x = \frac{k}{x} \)[/tex]

#### Step-by-Step Solution:
1. Multiply both sides by [tex]\( x \)[/tex] to eliminate the fraction:
[tex]\[ ax \cdot x = k \][/tex]
which simplifies to:
[tex]\[ ax^2 = k \][/tex]

2. Rewrite the equation in the standard quadratic form:
[tex]\[ ax^2 - k = 0 \][/tex]

3. Determine the number of solutions based on the discriminant of the quadratic equation. Recall that for a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], the discriminant ([tex]\(\Delta\)[/tex]) is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
In our equation [tex]\( ax^2 - k = 0 \)[/tex], the coefficients are:
[tex]\[ a = a, \quad b = 0, \quad c = -k \][/tex]

4. Calculate the discriminant:
[tex]\[ \Delta = 0^2 - 4 \cdot a \cdot (-k) = 4ak \][/tex]

5. Analyze the discriminant to find out the number of solutions:
- If [tex]\( 4ak > 0 \)[/tex], the equation [tex]\( ax^2 - k = 0 \)[/tex] has 2 real solutions.
- If [tex]\( 4ak = 0 \)[/tex], the equation [tex]\( ax^2 - k = 0 \)[/tex] has 1 real solution.
- If [tex]\( 4ak < 0 \)[/tex], the equation [tex]\( ax^2 - k = 0 \)[/tex] has no real solutions.

### Summary for [tex]\( a x = \frac{k}{x} \)[/tex]:
- 2 solutions if [tex]\( 4ak > 0 \)[/tex]
- 1 solution if [tex]\( 4ak = 0 \)[/tex]
- No solutions if [tex]\( 4ak < 0 \)[/tex]

### 2. Equation of the Form [tex]\( a x + b = \frac{1}{x} \)[/tex]

#### Step-by-Step Solution:
1. Multiply both sides by [tex]\( x \)[/tex] to eliminate the fraction:
[tex]\[ a x \cdot x + b \cdot x = 1 \][/tex]
which simplifies to:
[tex]\[ ax^2 + bx = 1 \][/tex]

2. Rewrite the equation in the standard quadratic form:
[tex]\[ ax^2 + bx - 1 = 0 \][/tex]

3. Determine the number of solutions based on the discriminant of the quadratic equation. Again, for a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], the discriminant ([tex]\(\Delta\)[/tex]) is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
In our equation [tex]\( ax^2 + bx - 1 = 0 \)[/tex], the coefficients are:
[tex]\[ a = a, \quad b = b, \quad c = -1 \][/tex]

4. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4 \cdot a \cdot (-1) = b^2 + 4a \][/tex]

5. Analyze the discriminant to determine the number of solutions:
- If [tex]\( b^2 + 4a > 0 \)[/tex], the equation [tex]\( ax^2 + bx - 1 = 0 \)[/tex] has 2 real solutions.
- If [tex]\( b^2 + 4a = 0 \)[/tex], the equation [tex]\( ax^2 + bx - 1 = 0 \)[/tex] has 1 real solution.
- If [tex]\( b^2 + 4a < 0 \)[/tex], the equation [tex]\( ax^2 + bx - 1 = 0 \)[/tex] has no real solutions.

### Summary for [tex]\( a x + b = \frac{1}{x} \)[/tex]:
- 2 solutions if [tex]\( b^2 + 4a > 0 \)[/tex]
- 1 solution if [tex]\( b^2 + 4a = 0 \)[/tex]
- No solutions if [tex]\( b^2 + 4a < 0 \)[/tex]

### Relationship to Quadratic Equations:
Both given equations can be transformed into quadratic equations through multiplication by [tex]\( x \)[/tex] and simplifying. The discriminant of these quadratic equations determines the number of real solutions each has. This is why studying the discriminant is crucial in solving these types of problems.
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