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Sagot :
To solve this problem, we need to find the value of [tex]\( x \)[/tex] given the dimensions of the rectangle and its area.
1. Identify the given information:
- The length of the rectangle is [tex]\( x \)[/tex] units.
- The width of the rectangle is [tex]\( (x - 15) \)[/tex] units.
- The area of the rectangle is 76 square units.
2. Write the formula for the area of a rectangle:
The area [tex]\( A \)[/tex] of a rectangle is given by the product of its length and width.
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ 76 = x \times (x - 15) \][/tex]
4. Set up the equation:
[tex]\[ x(x - 15) = 76 \][/tex]
5. Expand and simplify the equation:
[tex]\[ x^2 - 15x = 76 \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ x^2 - 15x - 76 = 0 \][/tex]
6. Solve the quadratic equation:
In order to solve the quadratic equation [tex]\( x^2 - 15x - 76 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -76 \)[/tex].
7. Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-15)^2 - 4(1)(-76) = 225 + 304 = 529 \][/tex]
8. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-(-15) \pm \sqrt{529}}{2(1)} = \frac{15 \pm 23}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{15 + 23}{2} = 19 \quad \text{and} \quad x = \frac{15 - 23}{2} = -4 \][/tex]
9. Select the valid solution:
Since a length cannot be negative, [tex]\( x = -4 \)[/tex] is not a valid solution. Therefore, the only valid solution is [tex]\( x = 19 \)[/tex].
10. Conclusion:
The value of [tex]\( x \)[/tex] is [tex]\( 19 \)[/tex].
Thus, the correct answer is:
[tex]\[ \boxed{19} \][/tex]
1. Identify the given information:
- The length of the rectangle is [tex]\( x \)[/tex] units.
- The width of the rectangle is [tex]\( (x - 15) \)[/tex] units.
- The area of the rectangle is 76 square units.
2. Write the formula for the area of a rectangle:
The area [tex]\( A \)[/tex] of a rectangle is given by the product of its length and width.
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ 76 = x \times (x - 15) \][/tex]
4. Set up the equation:
[tex]\[ x(x - 15) = 76 \][/tex]
5. Expand and simplify the equation:
[tex]\[ x^2 - 15x = 76 \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ x^2 - 15x - 76 = 0 \][/tex]
6. Solve the quadratic equation:
In order to solve the quadratic equation [tex]\( x^2 - 15x - 76 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -76 \)[/tex].
7. Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-15)^2 - 4(1)(-76) = 225 + 304 = 529 \][/tex]
8. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-(-15) \pm \sqrt{529}}{2(1)} = \frac{15 \pm 23}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{15 + 23}{2} = 19 \quad \text{and} \quad x = \frac{15 - 23}{2} = -4 \][/tex]
9. Select the valid solution:
Since a length cannot be negative, [tex]\( x = -4 \)[/tex] is not a valid solution. Therefore, the only valid solution is [tex]\( x = 19 \)[/tex].
10. Conclusion:
The value of [tex]\( x \)[/tex] is [tex]\( 19 \)[/tex].
Thus, the correct answer is:
[tex]\[ \boxed{19} \][/tex]
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