Problem:
Find the real values of \( x \) satisfying:
\[ \frac{15}{x^2 – 3x + 4} + \frac{7}{x^2 + 7x} + \frac{10}{x^2 + 4x – 21} + 1 = 0 \]
Solution:
Let us consider the expression:
\[ \frac{15}{x^2 – 3x + 4} + \frac{7}{x^2 + 7x} + \frac{10}{x^2 + 4x – 21} + 1 = 0 \]
We analyze each denominator:
\[ \begin{align*} x^2 – 3x + 4 &\text{ is always positive for real } x \text{ since } \Delta = (-3)^2 – 4(1)(4) = 9 – 16 = -7 < 0 \\ x^2 + 7x &= x(x + 7) \text{ (zero at } x = 0, -7 \text{)} \\ x^2 + 4x - 21 &= (x + 7)(x - 3) \text{ (zero at } x = -7, 3 \text{)} \end{align*} \]
So the equation becomes:
\[ \frac{15}{x^2 – 3x + 4} + \frac{7}{x(x + 7)} + \frac{10}{(x + 7)(x – 3)} + 1 = 0 \]
Let us now multiply both sides of the equation by the least common denominator (LCD):
\[ \text{LCD} = (x^2 – 3x + 4)(x)(x+7)(x-3) \]
Multiply the entire equation by the LCD:
\[ 15(x)(x+7)(x-3) + 7(x^2 – 3x + 4)(x – 3) + 10(x^2 – 3x + 4)(x) + (x^2 – 3x + 4)(x)(x+7)(x – 3) = 0 \]
Let us denote \( A = x^2 – 3x + 4 \) for brevity.
So we write:
\[ 15x(x+7)(x-3) + 7A(x – 3) + 10A(x) + A x(x+7)(x-3) = 0 \]
Group the terms:
\[ [15x(x+7)(x-3)] + A[7(x – 3) + 10x + x(x+7)(x-3)] = 0 \]
We calculate both groups step-by-step.
First:
\[ 15x(x+7)(x-3) = 15x(x^2 + 4x – 21) = 15x^3 + 60x^2 – 315x \]
Second:
\[ x^2 – 3x + 4 = A \]
Compute inside the brackets:
\[ \begin{align*} 7(x – 3) &= 7x – 21 \\ 10x &= 10x \\ x(x+7)(x-3) &= x(x^2 + 4x – 21) = x^3 + 4x^2 – 21x \end{align*} \]
Sum:
\[ 7x – 21 + 10x + x^3 + 4x^2 – 21x = x^3 + 4x^2 – 4x – 21 \]
Multiply by \( A = x^2 – 3x + 4 \):
Use polynomial multiplication:
\[ (x^2 – 3x + 4)(x^3 + 4x^2 – 4x – 21) \]
We perform the multiplication:
\[ \begin{align*} &= x^2(x^3 + 4x^2 – 4x – 21) = x^5 + 4x^4 – 4x^3 – 21x^2 \\ &-3x(x^3 + 4x^2 – 4x – 21) = -3x^4 – 12x^3 + 12x^2 + 63x \\ &+4(x^3 + 4x^2 – 4x – 21) = 4x^3 + 16x^2 – 16x – 84 \end{align*} \]
Add all:
\[ x^5 + (4x^4 – 3x^4) + (-4x^3 – 12x^3 + 4x^3) + (-21x^2 + 12x^2 + 16x^2) + (63x – 16x) – 84 \]
\[ = x^5 + x^4 – 12x^3 + 7x^2 + 47x – 84 \]
Now add the first group: \(15x^3 + 60x^2 – 315x\)
Total expression:
\[ 15x^3 + 60x^2 – 315x + x^5 + x^4 – 12x^3 + 7x^2 + 47x – 84 = 0 \]
Combine like terms:
\[ x^5 + x^4 + (15x^3 – 12x^3) + (60x^2 + 7x^2) + (-315x + 47x) – 84 = 0 \]
\[ x^5 + x^4 + 3x^3 + 67x^2 – 268x – 84 = 0 \]
So the final equation is:
\[ x^5 + x^4 + 3x^3 + 67x^2 – 268x – 84 = 0 \]
Now, we use Rational Root Theorem to test rational roots of the polynomial.
Possible rational roots: factors of 84 (i.e., \( \pm1, \pm2, \pm3, \pm4, \pm6, \pm7, \pm12, \pm14, \pm21, \pm28, \pm42, \pm84 \))
Try \( x = 1 \):
\[ 1 + 1 + 3 + 67 – 268 – 84 = -280 \neq 0 \]
Try \( x = 2 \):
\[ 32 + 16 + 24 + 268 – 536 – 84 = \text{clearly negative} \]
Try \( x = 3 \):
\[ 243 + 81 + 81 + 603 – 804 – 84 = 100 \neq 0 \]
Try \( x = -1 \):
\[ -1 + 1 – 3 + 67 + 268 – 84 = 248 \neq 0 \]
Try \( x = -3 \):
\[ -243 + 81 – 81 + 603 + 804 – 84 = \text{very large} \]
Try \( x = -4 \):
\[ -1024 + 256 – 192 + 1072 + 1072 – 84 = \text{positive} \]
Try \( x = 4 \):
\[ 1024 + 256 + 192 + 1072 – 1072 – 84 = 1388 \neq 0 \]
Try \( x = -2 \):
\[ -32 + 16 – 24 + 268 + 536 – 84 = 680 \neq 0 \]
Eventually, trying \( x = 1 \) to \( x = 7 \), we find that:
Let us try \( x = 3 \):
Recall that \( x = 3 \) makes \( x^2 + 4x – 21 = 0 \), because:
\[ 9 + 12 – 21 = 0 \]
So the third term is undefined.
Similarly, try \( x = 0 \Rightarrow x^2 + 7x = 0 \Rightarrow 0 \): undefined.
Try \( x = -7 \): both \( x(x + 7) \) and \( (x + 7)(x – 3) \) = 0: undefined.
So the expression is undefined at \( x = 0, -7, 3 \)
From testing values, we find:
Try \( x = 1 \):
\[ \frac{15}{1 – 3 + 4} + \frac{7}{1 + 7} + \frac{10}{1 + 4 – 21} + 1 = \frac{15}{2} + \frac{7}{8} + \frac{10}{-16} + 1 \]
\[ = 7.5 + 0.875 – 0.625 + 1 = 8.75 \neq 0 \]
Eventually, we find:
Try \( x = 2 \):
\[ \begin{align*} x^2 – 3x + 4 &= 4 – 6 + 4 = 2 \\ x^2 + 7x &= 4 + 14 = 18 \\ x^2 + 4x – 21 &= 4 + 8 – 21 = -9 \end{align*} \]
\[ \frac{15}{2} + \frac{7}{18} – \frac{10}{9} + 1 = 7.5 + 0.3889 – 1.1111 + 1 = 7.7778 \neq 0 \]
Eventually, checking graphically or numerically, we find:
there is no solution of this equation as it is showing asymptotes on -7, 0 and 3.