Question

3х² + 1 : 2х = х + 1

Answer 1

3х² + 1 : 2х = х + 1

3х² + 1/2х = х + 1

3х² + 1/2х - х - 1 = 0

3х² - 1/2х - 1 = 0

3х² - 0,5х - 1 = 0

D = (- 0,5)² - 4 * 3 * (-1) = 12,25

        0,5 - √12,25

х₁ = ----------------------- = - 3/6 = - 1/2

            2 * 3

        0,5 + √12,25

х₂ = ----------------------- = 2/3

              2 * 3

Answer 2

To solve the given equation, we'll first need to simplify it and then find the value of 'x' that satisfies the equation. Here's the step-by-step solution:

Step 1: Simplify the equation: $\frac{3x^2 + 1}{2x} = x + 1$

Step 2: Eliminate the fraction by multiplying both sides of the equation by 2x (the denominator on the left side): $2x \times \frac{3x^2 + 1}{2x} = 2x \times (x + 1)$

On the left side, the 2x in the numerator cancels out with the 2x in the denominator, leaving us with: $3x^2 + 1 = 2x \times (x + 1)$

Step 3: Distribute the 2x on the right side: $3x^2 + 1 = 2x^2 + 2x$

Step 4: Move all the terms to one side of the equation to set it to zero: $3x^2 + 1 - (2x^2 + 2x) = 0$

Step 5: Combine like terms: $3x^2 - 2x^2 + 2x - 1 = 0$

Step 6: Simplify the equation further: $x^2 + 2x - 1 = 0$

Step 7: Now, we need to solve this quadratic equation for 'x'. We can do this by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula states that for an equation of the form: $ax^2 + bx + c = 0$, the solutions for 'x' are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 1$, $b = 2$, and $c = -1$. Plugging these values into the formula, we get: $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}$ $x = \frac{-2 \pm \sqrt{4 + 4}}{2}$ $x = \frac{-2 \pm \sqrt{8}}{2}$ $x = \frac{-2 \pm 2\sqrt{2}}{2}$

Step 8: Simplify further: $x = \frac{-2 + 2\sqrt{2}}{2} \quad \text{or} \quad x = \frac{-2 - 2\sqrt{2}}{2}$

Step 9: Finally, simplify each solution: $x = -1 + \sqrt{2} \quad \text{or} \quad x = -1 - \sqrt{2}$

So, the two solutions to the equation are approximately: $x \approx 0.414 \quad \text{or} \quad x \approx -2.414$