the length of a rectangle exceeds its width by 8cm and the area of the rectangle is 240 cm 2 represent the above situation in the form of a quadratic equation

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Solution

Let the breadth of the rectangle = x cm
length of the rectangle = (x + 8) cm
Area of rectangle = length Ã— breadth = x (x + 8) = x²+ 8x
but area of rectangle = 240 cm²
according to the given condition,
x²+ 8x = 240
x²+ 8x - 240 = 0

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the length of a rectangle exceeds its width by 8cm and the area of the rectangle is 240 cm 2 represent the above situation in the form of a quadratic equation

Let the breadth of the rectangle = x cm length of the rectangle = (x + 8) cmArea of rectangle = length Ã— breadth = x (x + 8) = x2 + 8xbut area of rectangle = 240 cm2according to the given condition, x2 + 8x = 240 x2 + 8x - 240 = 0

Let the breadth of the rectangle = x cm
length of the rectangle = (x + 8) cm
Area of rectangle = length Ã— breadth = x (x + 8) = x²+ 8x
but area of rectangle = 240 cm²
according to the given condition,
x²+ 8x = 240
x²+ 8x - 240 = 0