The length of a rectangle is 4 cm more than 3 times its width. If the area of the rectangle is 15 cm, which equation could be used to find the width, w?
w = width and length=l = (3w+4)
area = w x l
area = 15
15 = w * (3w+4)
Area = L x W
Let the width of the rectangle =W
Then the length is =3W + 4
15 =W x [3W + 4], solve for W
Solve for W:
15 = W (3 W + 4)
15 = W (3 W + 4) is equivalent to W (3 W + 4) = 15:
W (3 W + 4) = 15
Expand out terms of the left hand side:
3 W^2 + 4 W = 15
Divide both sides by 3:
W^2 + (4 W)/3 = 5
Add 4/9 to both sides:
W^2 + (4 W)/3 + 4/9 = 49/9
Write the left hand side as a square:
(W + 2/3)^2 = 49/9
Take the square root of both sides:
W + 2/3 = 7/3 or W + 2/3 = -7/3
Subtract 2/3 from both sides:
W = 5/3 or W + 2/3 = -7/3
Subtract 2/3 from both sides:
Answer: |W = 5/3 or W = -3 Discard
w = width and length=l = (3w+4)
area = w x l
area = 15
15 = w * (3w+4)
The length of a rectangle is 4 cm more than 3 times its width. If the area of the rectangle is 15 cm^2, which equation could be used to find the width, w?
\(l=3w+4cm\)
\(15cm^2=(3w+4cm)\times w\)
\(15cm^2=3w^2+4cm\times w\)
\(3w^2+4cm\times w-15cm^2=0\) \([w = {-b + \sqrt{b^2-4ac} \over 2a}]\)
a b c
\(\large w=\frac{-4cm+\sqrt{16cm^2+180cm^2}}{6}\)
\(w=\frac{-4cm+{14cm}}{6}=\frac{10cm}{6}\)
\(\large w=1\frac{2}{3}cm\)
\(l=3w+4cm=9cm\)
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