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Final: The Rug, Solved

Part of the Factoring & Quadratics section of Coddy's Algebra journey. Lesson 60 of 61.

To solve an equation from a word problem, first expand any product. Then move every term to one side so that the other side is zero, factor, and set each factor equal to zero. Finally, check each solution against the story. A width or a length must be positive, so a negative solution is rejected.

For example, a rectangle with width xx, length x+2x+2 and area 15 gives x(x+2)=15x\left(x+2\right) = 15. Expanding gives x2+2x=15x^{2}+2x = 15, and subtracting 15 from both sides gives x2+2x−15=0x^{2}+2x-15 = 0. This factors as (x+5)(x−3)=0\left(x+5\right)\left(x-3\right) = 0, so x=−5x = -5 or x=3x = 3. A width cannot be -5, so the width is 3 and the length is 5. Both values solve the equation, but only one of them fits the rectangle.

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Challenge

Hard

Solve x(x+3)=40x\left(x+3\right) = 40 for xx. Give both solutions, then decide which one can be the width of the rug.

Try it yourself

x(x+3)=40x\left(x+3\right) = 40

Get xx alone

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