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Sagot :
Certainly! Let's work through the problem step-by-step to find the value of [tex]\( x \)[/tex].
Given the equation for the area of the trapezium:
[tex]\[ \frac{[(x+3)+(x-1)] \cdot x}{2} = 35.75 \][/tex]
1. Combine the terms in the numerator:
[tex]\[(x+3)+(x-1) = x + x + 3 - 1 = 2x + 2\][/tex]
2. Substitute this back into the equation:
[tex]\[ \frac{(2x + 2) \cdot x}{2} = 35.75 \][/tex]
3. Eliminate the fraction by multiplying both sides of the equation by 2:
[tex]\[ (2x + 2) \cdot x = 71.5 \][/tex]
4. Distribute [tex]\( x \)[/tex] on the left-hand side:
[tex]\[ 2x^2 + 2x = 71.5 \][/tex]
5. Rearrange the equation to set it to 0 (standard form of a quadratic equation):
[tex]\[ 2x^2 + 2x - 71.5 = 0 \][/tex]
6. Divide the entire equation by 2 to simplify:
[tex]\[ x^2 + x - 35.75 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -35.75 \)[/tex].
7. Solving the quadratic equation:
To find the roots of the quadratic equation [tex]\( x^2 + x - 35.75 = 0 \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -35.75 \)[/tex].
Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-35.75)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 143}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{144}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm 12}{2} \][/tex]
8. Calculate the two possible values for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-1 + 12}{2} = \frac{11}{2} = 5.5 \][/tex]
[tex]\[ x = \frac{-1 - 12}{2} = \frac{-13}{2} = -6.5 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] that satisfy the equation are:
[tex]\[ x = 5.5 \quad \text{and} \quad x = -6.5 \][/tex]
Given the context of the problem, where lengths typically cannot be negative, the most plausible value for [tex]\( x \)[/tex] is:
[tex]\[ x = 5.5 \][/tex]
However, mathematically, both [tex]\( 5.5 \)[/tex] and [tex]\( -6.5 \)[/tex] are solutions to the quadratic equation.
Given the equation for the area of the trapezium:
[tex]\[ \frac{[(x+3)+(x-1)] \cdot x}{2} = 35.75 \][/tex]
1. Combine the terms in the numerator:
[tex]\[(x+3)+(x-1) = x + x + 3 - 1 = 2x + 2\][/tex]
2. Substitute this back into the equation:
[tex]\[ \frac{(2x + 2) \cdot x}{2} = 35.75 \][/tex]
3. Eliminate the fraction by multiplying both sides of the equation by 2:
[tex]\[ (2x + 2) \cdot x = 71.5 \][/tex]
4. Distribute [tex]\( x \)[/tex] on the left-hand side:
[tex]\[ 2x^2 + 2x = 71.5 \][/tex]
5. Rearrange the equation to set it to 0 (standard form of a quadratic equation):
[tex]\[ 2x^2 + 2x - 71.5 = 0 \][/tex]
6. Divide the entire equation by 2 to simplify:
[tex]\[ x^2 + x - 35.75 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -35.75 \)[/tex].
7. Solving the quadratic equation:
To find the roots of the quadratic equation [tex]\( x^2 + x - 35.75 = 0 \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -35.75 \)[/tex].
Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-35.75)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 143}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{144}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm 12}{2} \][/tex]
8. Calculate the two possible values for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-1 + 12}{2} = \frac{11}{2} = 5.5 \][/tex]
[tex]\[ x = \frac{-1 - 12}{2} = \frac{-13}{2} = -6.5 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] that satisfy the equation are:
[tex]\[ x = 5.5 \quad \text{and} \quad x = -6.5 \][/tex]
Given the context of the problem, where lengths typically cannot be negative, the most plausible value for [tex]\( x \)[/tex] is:
[tex]\[ x = 5.5 \][/tex]
However, mathematically, both [tex]\( 5.5 \)[/tex] and [tex]\( -6.5 \)[/tex] are solutions to the quadratic equation.
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