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Sagot :
Let's solve the equation step-by-step:
Given equation:
[tex]\[ \frac{x^2 - x - 6}{x^2} = \frac{x - 6}{2x} + \frac{2x + 12}{x} \][/tex]
Step 1: Identify the least common denominator (LCD) for the fractions.
The denominators are [tex]\(x^2\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x\)[/tex]. The least common denominator is [tex]\(2x^2\)[/tex].
Step 2: Multiply each term on both sides of the equation by the LCD ([tex]\(2x^2\)[/tex]) to clear the fractions.
Multiplying each side:
[tex]\[ 2x^2 \cdot \frac{x^2 - x - 6}{x^2} = 2x^2 \cdot \left(\frac{x - 6}{2x} + \frac{2x + 12}{x}\right) \][/tex]
This simplifies to:
[tex]\[ 2(x^2 - x - 6) = 2x^2 \cdot \frac{x - 6}{2x} + 2x^2 \cdot \frac{2x + 12}{x} \][/tex]
Step 3: Simplify each term.
Starting with the left-hand side:
[tex]\[ 2(x^2 - x - 6) = 2x^2 - 2x - 12 \][/tex]
On the right-hand side, simplify each multiplication:
[tex]\[ 2x^2 \cdot \frac{x - 6}{2x} = x(x - 6) = x^2 - 6x \][/tex]
[tex]\[ 2x^2 \cdot \frac{2x + 12}{x} = 2x(2x + 12) = 4x^2 + 24x \][/tex]
Adding these together:
[tex]\[ x^2 - 6x + 4x^2 + 24x = 5x^2 + 18x \][/tex]
Now the equation is:
[tex]\[ 2x^2 - 2x - 12 = 5x^2 + 18x \][/tex]
Step 4: Move all terms to one side to set the equation to zero:
[tex]\[ 2x^2 - 2x - 12 - 5x^2 - 18x = 0 \][/tex]
Combine like terms:
[tex]\[ -3x^2 - 20x - 12 = 0 \][/tex]
Step 5: To find the solutions for [tex]\(x\)[/tex], solve the quadratic equation [tex]\(-3x^2 - 20x - 12 = 0\)[/tex].
This quadratic equation can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = -3\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = -12\)[/tex].
First, calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-20)^2 - 4(-3)(-12) = 400 - 144 = 256 \][/tex]
Since the discriminant is positive, we will have two real solutions:
[tex]\[ x = \frac{-(-20) \pm \sqrt{256}}{2(-3)} = \frac{20 \pm 16}{-6} \][/tex]
Now calculate the two solutions:
[tex]\[ x = \frac{20 + 16}{-6} = \frac{36}{-6} = -6 \][/tex]
[tex]\[ x = \frac{20 - 16}{-6} = \frac{4}{-6} = -\frac{2}{3} \][/tex]
Thus, the solutions are:
[tex]\[ x = -6 \quad \text{and} \quad x = -\frac{2}{3} \][/tex]
Finally, we should verify whether these values do not make any of the denominators in the original equation zero.
For [tex]\(x = -6\)[/tex]: The denominators [tex]\(x^2\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x\)[/tex] are not zero.
For [tex]\(x = -\frac{2}{3}\)[/tex]: The denominators [tex]\(x^2\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x\)[/tex] are not zero.
Thus, the solutions [tex]\(x = -6\)[/tex] and [tex]\(x = -\frac{2}{3}\)[/tex] are valid solutions to the equation.
Given equation:
[tex]\[ \frac{x^2 - x - 6}{x^2} = \frac{x - 6}{2x} + \frac{2x + 12}{x} \][/tex]
Step 1: Identify the least common denominator (LCD) for the fractions.
The denominators are [tex]\(x^2\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x\)[/tex]. The least common denominator is [tex]\(2x^2\)[/tex].
Step 2: Multiply each term on both sides of the equation by the LCD ([tex]\(2x^2\)[/tex]) to clear the fractions.
Multiplying each side:
[tex]\[ 2x^2 \cdot \frac{x^2 - x - 6}{x^2} = 2x^2 \cdot \left(\frac{x - 6}{2x} + \frac{2x + 12}{x}\right) \][/tex]
This simplifies to:
[tex]\[ 2(x^2 - x - 6) = 2x^2 \cdot \frac{x - 6}{2x} + 2x^2 \cdot \frac{2x + 12}{x} \][/tex]
Step 3: Simplify each term.
Starting with the left-hand side:
[tex]\[ 2(x^2 - x - 6) = 2x^2 - 2x - 12 \][/tex]
On the right-hand side, simplify each multiplication:
[tex]\[ 2x^2 \cdot \frac{x - 6}{2x} = x(x - 6) = x^2 - 6x \][/tex]
[tex]\[ 2x^2 \cdot \frac{2x + 12}{x} = 2x(2x + 12) = 4x^2 + 24x \][/tex]
Adding these together:
[tex]\[ x^2 - 6x + 4x^2 + 24x = 5x^2 + 18x \][/tex]
Now the equation is:
[tex]\[ 2x^2 - 2x - 12 = 5x^2 + 18x \][/tex]
Step 4: Move all terms to one side to set the equation to zero:
[tex]\[ 2x^2 - 2x - 12 - 5x^2 - 18x = 0 \][/tex]
Combine like terms:
[tex]\[ -3x^2 - 20x - 12 = 0 \][/tex]
Step 5: To find the solutions for [tex]\(x\)[/tex], solve the quadratic equation [tex]\(-3x^2 - 20x - 12 = 0\)[/tex].
This quadratic equation can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = -3\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = -12\)[/tex].
First, calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-20)^2 - 4(-3)(-12) = 400 - 144 = 256 \][/tex]
Since the discriminant is positive, we will have two real solutions:
[tex]\[ x = \frac{-(-20) \pm \sqrt{256}}{2(-3)} = \frac{20 \pm 16}{-6} \][/tex]
Now calculate the two solutions:
[tex]\[ x = \frac{20 + 16}{-6} = \frac{36}{-6} = -6 \][/tex]
[tex]\[ x = \frac{20 - 16}{-6} = \frac{4}{-6} = -\frac{2}{3} \][/tex]
Thus, the solutions are:
[tex]\[ x = -6 \quad \text{and} \quad x = -\frac{2}{3} \][/tex]
Finally, we should verify whether these values do not make any of the denominators in the original equation zero.
For [tex]\(x = -6\)[/tex]: The denominators [tex]\(x^2\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x\)[/tex] are not zero.
For [tex]\(x = -\frac{2}{3}\)[/tex]: The denominators [tex]\(x^2\)[/tex], [tex]\(2x\)[/tex], and [tex]\(x\)[/tex] are not zero.
Thus, the solutions [tex]\(x = -6\)[/tex] and [tex]\(x = -\frac{2}{3}\)[/tex] are valid solutions to the equation.
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