_{the length of a rectangle is 3cm more than the width the area is 70cm2}

^{2 }are both given, the equation then becomes

^{2}+3W by distribution.

^{2 }+ 3W - 70 = 0

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Area is equal to the length times the width. Thus, A = L·W. Since L = W + 3 and A = 70 cm^{2 }are both given, the equation then becomes

70 = (W+3)W

70 = W^{2}+3W by distribution.

Next, set the resulting quadratic equation equal to zero, factor, then use the zero-product property to solve for W.

W^{2 }+ 3W - 70 = 0

(W+10)(W - 7) = 0

W = -10, 7.

Since it does not make sense that the width is negative, W must be 7 cm. We were given the length of the rectangle was 3 more than the width, so the length, L, must be 10 cm.

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor

Let L = the length of the rectangle and W = its width

Area = L*W = 70 cm^{2} [The Area is 70 cm2]

L = W+3 [The length of a rectangle is 3 cm more than its width]

Substitute W+3 in place of L in the Area equation:

Area = 70 cm^{2} = (W+3)*W

70 = W^{2} + 3W

0 = W^{2} + 3W - 70

Factors to:

(W+10)(W-7)

W = -10 and 7

Can't have a negative length, so **W = 7**

Solve for L:

L = W + 3

L= 7 + 3

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