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A sand box has an area of 45 ft. The length is 4 feet longer than the width. What are the dimensions of the sand box? Solve by completing the square

Sagot :

SchwarzschildRadius
So,

The sand box's area is 45 ft.

The length is 4 feet longer than the width.

We can solve by translating these sentences into mathematical form.

Let l represent length and w represent width.

First equation: lw = 45

Second equation: l = 4 + w

Substitute 4 + w for l in the first equation.
(4 + w)w = 45

Distribute
[tex] w^{2} + 4w = 45[/tex]

Subtract 45 from both sides
[tex] w^{2} + 4w - 45[/tex]

Factor
[tex](w + 9)(w -5) = w^{2} + 4w - 45[/tex]

Set both factors equal to zero.
w + 9 = 0
w - 5 = 0

Subtract 9 from both sides
w = -9

Add 5 to both sides
w = 5

We have to cross out -9 for the width, because, logically, it is impossible to have a negative width.

Substitute 5 for w in the second original equation.
l = 4 + (5)
l = 9

Check
5 * 9 = 45
45 = 45 This checks.

9 = 4 + 5
9 = 9 This also checks.

The length is 9 ft. and the width is 4 ft.
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