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The length of a rectangle is 17 inches longer than the width (x),Which is the width (x) when the area (y) is 1334 square inches?

The Length Of A Rectangle Is 17 Inches Longer Than The Width XWhich Is The Width X When The Area Y Is 1334 Square Inches class=

Sagot :

x: width

z: length

The length of a rectangle is 17 inches longer than the width (x) means:

z = x + 17

The area of a rectangle is computed as follows:

Area = x*z

Replacing with Area = 1334 and z = x+17:

1334 = x*(x + 17)

Applying distributive property:

1334 = x*x + x*17

0 = x² + 17x - 1334

Using the quadratic formula:

[tex]\begin{gathered} x_{1,2}=\frac{-17\pm\sqrt[]{17^2-4\cdot1\cdot(-1334)}}{2\cdot1} \\ x_{1,2}=\frac{-17\pm\sqrt[]{5625}}{2} \\ x_1=\frac{-17+75}{2}=29_{} \\ x_2=\frac{-17-75}{2}=-46_{} \end{gathered}[/tex]

Given that x cannot be negative, then the answer is x = 29 inches

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