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Sagot :
To solve this rational equation and determine which value must be discarded due to the context of the problem, let's analyze the situation step-by-step.
Given the equation:
[tex]\[ \frac{1}{x} + \frac{1}{x+18} = \frac{1}{40} \][/tex]
we are representing the scenario where [tex]\(x\)[/tex] is the number of minutes it takes Niki to paint the room alone, and [tex]\(x+18\)[/tex] is the time it takes Greg to paint the room alone.
First, identify the common denominators:
The least common denominator for [tex]\(x\)[/tex], [tex]\(x+18\)[/tex], and [tex]\(40\)[/tex] is [tex]\(40x(x+18)\)[/tex]. Multiply every term in the equation by [tex]\(40x(x+18)\)[/tex] to clear the fractions:
[tex]\[ 40x(x+18) \left(\frac{1}{x}\right) + 40x(x+18) \left(\frac{1}{x+18}\right) = 40x(x+18) \left(\frac{1}{40}\right) \][/tex]
Simplify each term:
[tex]\[ 40(x+18) + 40x = x(x+18) \][/tex]
Now expand and combine like terms:
[tex]\[ 40x + 720 + 40x = x^2 + 18x \][/tex]
Combine all terms:
[tex]\[ 80x + 720 = x^2 + 18x \][/tex]
Subtract [tex]\(80x + 720\)[/tex] from both sides to set the equation to zero:
[tex]\[ x^2 + 18x - 80x - 720 = 0 \][/tex]
[tex]\[ x^2 - 62x - 720 = 0 \][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], where:
[tex]\[ a = 1, \, b = -62, \, c = -720 \][/tex]
To solve for [tex]\(x\)[/tex], use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x = \frac{-(-62) \pm \sqrt{(-62)^2 - 4(1)(-720)}}{2(1)} \][/tex]
[tex]\[ x = \frac{62 \pm \sqrt{3844 + 2880}}{2} \][/tex]
[tex]\[ x = \frac{62 \pm \sqrt{6724}}{2} \][/tex]
[tex]\[ x = \frac{62 \pm 82}{2} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{62 + 82}{2} = \frac{144}{2} = 72 \][/tex]
[tex]\[ x = \frac{62 - 82}{2} = \frac{-20}{2} = -10 \][/tex]
Since [tex]\(x\)[/tex] represents the number of minutes it takes Niki to paint the room, a negative value for [tex]\(x\)[/tex] does not make sense in this context.
Now, consider the given options:
A. [tex]\(x = 0\)[/tex]
B. [tex]\(x = -10\)[/tex]
C. [tex]\(x = -18\)[/tex]
D. [tex]\(x = -72\)[/tex]
Each of the given values is either zero or negative, and must be discarded because they do not fit the physical context of the problem (time taken to paint must be positive). Thus, the values that must be discarded in this context are:
[tex]\[ \boxed{0, -10, -18, -72} \][/tex]
Given the equation:
[tex]\[ \frac{1}{x} + \frac{1}{x+18} = \frac{1}{40} \][/tex]
we are representing the scenario where [tex]\(x\)[/tex] is the number of minutes it takes Niki to paint the room alone, and [tex]\(x+18\)[/tex] is the time it takes Greg to paint the room alone.
First, identify the common denominators:
The least common denominator for [tex]\(x\)[/tex], [tex]\(x+18\)[/tex], and [tex]\(40\)[/tex] is [tex]\(40x(x+18)\)[/tex]. Multiply every term in the equation by [tex]\(40x(x+18)\)[/tex] to clear the fractions:
[tex]\[ 40x(x+18) \left(\frac{1}{x}\right) + 40x(x+18) \left(\frac{1}{x+18}\right) = 40x(x+18) \left(\frac{1}{40}\right) \][/tex]
Simplify each term:
[tex]\[ 40(x+18) + 40x = x(x+18) \][/tex]
Now expand and combine like terms:
[tex]\[ 40x + 720 + 40x = x^2 + 18x \][/tex]
Combine all terms:
[tex]\[ 80x + 720 = x^2 + 18x \][/tex]
Subtract [tex]\(80x + 720\)[/tex] from both sides to set the equation to zero:
[tex]\[ x^2 + 18x - 80x - 720 = 0 \][/tex]
[tex]\[ x^2 - 62x - 720 = 0 \][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], where:
[tex]\[ a = 1, \, b = -62, \, c = -720 \][/tex]
To solve for [tex]\(x\)[/tex], use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ x = \frac{-(-62) \pm \sqrt{(-62)^2 - 4(1)(-720)}}{2(1)} \][/tex]
[tex]\[ x = \frac{62 \pm \sqrt{3844 + 2880}}{2} \][/tex]
[tex]\[ x = \frac{62 \pm \sqrt{6724}}{2} \][/tex]
[tex]\[ x = \frac{62 \pm 82}{2} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{62 + 82}{2} = \frac{144}{2} = 72 \][/tex]
[tex]\[ x = \frac{62 - 82}{2} = \frac{-20}{2} = -10 \][/tex]
Since [tex]\(x\)[/tex] represents the number of minutes it takes Niki to paint the room, a negative value for [tex]\(x\)[/tex] does not make sense in this context.
Now, consider the given options:
A. [tex]\(x = 0\)[/tex]
B. [tex]\(x = -10\)[/tex]
C. [tex]\(x = -18\)[/tex]
D. [tex]\(x = -72\)[/tex]
Each of the given values is either zero or negative, and must be discarded because they do not fit the physical context of the problem (time taken to paint must be positive). Thus, the values that must be discarded in this context are:
[tex]\[ \boxed{0, -10, -18, -72} \][/tex]
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