Question

The number of real roots of the equation $\frac{4x}{x^2+x+3} + \frac{5x}{x^2-5x+3} = \frac{-3}{2}$ is equal to

Answer 1

2

Answer 2

The given equation is: $\frac{4x}{x^2+x+3} + \frac{5x}{x^2-5x+3} = \frac{-3}{2}$

Step 1: Check for $x=0$.

If $x=0$, the equation becomes $\frac{0}{3} + \frac{0}{3} = 0$. The right-hand side is $\frac{-3}{2}$. Since $0 \neq \frac{-3}{2}$, $x=0$ is not a root of the equation. This means we can safely divide by $x$ without losing any solutions.

Step 2: Divide numerator and denominator by $x$.

Since $x \neq 0$, we can divide the numerator and denominator of each term on the left-hand side by $x$: $\frac{\frac{4x}{x}}{\frac{x^2+x+3}{x}} + \frac{\frac{5x}{x}}{\frac{x^2-5x+3}{x}} = \frac{-3}{2}$ $\frac{4}{x+1+\frac{3}{x}} + \frac{5}{x-5+\frac{3}{x}} = \frac{-3}{2}$

Step 3: Introduce a substitution.

Let $y = x + \frac{3}{x}$. Substituting this into the equation: $\frac{4}{y+1} + \frac{5}{y-5} = \frac{-3}{2}$

Step 4: Solve the equation for $y$.

Combine the fractions on the left-hand side: $\frac{4(y-5) + 5(y+1)}{(y+1)(y-5)} = \frac{-3}{2}$ $\frac{4y - 20 + 5y + 5}{y^2 - 5y + y - 5} = \frac{-3}{2}$ $\frac{9y - 15}{y^2 - 4y - 5} = \frac{-3}{2}$

Cross-multiply: $2(9y - 15) = -3(y^2 - 4y - 5)$ $18y - 30 = -3y^2 + 12y + 15$

Move all terms to one side to form a quadratic equation in $y$: $3y^2 + 18y - 12y - 30 - 15 = 0$ $3y^2 + 6y - 45 = 0$

Divide the entire equation by 3: $y^2 + 2y - 15 = 0$

Factor the quadratic equation: $(y+5)(y-3) = 0$

This gives two possible values for $y$: $y = -5$ or $y = 3$

Step 5: Substitute back and solve for $x$.

Case 1: $y = -5$

Substitute $y = x + \frac{3}{x}$: $x + \frac{3}{x} = -5$ Multiply by $x$ (since $x \neq 0$): $x^2 + 3 = -5x$ $x^2 + 5x + 3 = 0$

This is a quadratic equation in $x$. Let's find its discriminant, $\Delta_1 = b^2 - 4ac$: $\Delta_1 = (5)^2 - 4(1)(3) = 25 - 12 = 13$ Since $\Delta_1 = 13 > 0$, this quadratic equation has two distinct real roots.

Case 2: $y = 3$

Substitute $y = x + \frac{3}{x}$: $x + \frac{3}{x} = 3$ Multiply by $x$: $x^2 + 3 = 3x$ $x^2 - 3x + 3 = 0$

This is a quadratic equation in $x$. Let's find its discriminant, $\Delta_2 = b^2 - 4ac$: $\Delta_2 = (-3)^2 - 4(1)(3) = 9 - 12 = -3$ Since $\Delta_2 = -3 < 0$, this quadratic equation has no real roots (it has two complex conjugate roots).

Step 6: Check for domain restrictions.

The original denominators are $x^2+x+3$ and $x^2-5x+3$. For $x^2+x+3=0$, the discriminant is $1^2 - 4(1)(3) = 1 - 12 = -11 < 0$. So, $x^2+x+3$ is never zero for real $x$. For $x^2-5x+3=0$, the discriminant is $(-5)^2 - 4(1)(3) = 25 - 12 = 13 > 0$. The roots are $x = \frac{5 \pm \sqrt{13}}{2}$. The real roots we found from $x^2+5x+3=0$ are $x = \frac{-5 \pm \sqrt{13}}{2}$. These are clearly different from $\frac{5 \pm \sqrt{13}}{2}$. Also, the values of $y$ we found ($y=-5$ and $y=3$) do not make the denominators $(y+1)$ or $(y-5)$ zero in the substituted equation. If $y=-1$, then $x^2+x+3=0$, which has no real roots. If $y=5$, then $x^2-5x+3=0$, which has real roots, but our $y$ values are not $5$. Thus, the two real roots obtained are valid.

Conclusion:

From Case 1, we found 2 distinct real roots. From Case 2, we found 0 real roots. Therefore, the total number of real roots for the given equation is $2 + 0 = 2$.