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I’m stuck on the specific question please help it #2

I’m stuck on the specific question please help it #2-example-1

1 Answer

6 votes

Let

x be the width

x + 25 be the length

Given that the area is 7154 square feet, then we can substitute the following using the formula for the area of a rectangle


\begin{gathered} A_{\text{rectangle}}=lw \\ 7154=(x+25)(x) \end{gathered}

Simplify the left side and equate to zero


\begin{gathered} 7154=(x+25)(x) \\ 7154=x^2+25x \\ x^2+25x-7154=0 \end{gathered}

Now it is in the standard quadratic equation with the following coefficients, a = 1, b = 25, and c = -7154.

Use the quadratic formula to solve for x


\begin{gathered} x = ( -b \pm √(b^2 - 4ac))/( 2a ) \\ x = ( -25 \pm √(25^2 - 4(1)(-7154)))/( 2(1) ) \\ x = ( -25 \pm √(625 - -28616))/( 2 ) \\ x=(-25\pm√(625+28616))/(2) \\ x = ( -25 \pm √(29241))/( 2 ) \\ x = ( -25 \pm 171\, )/( 2 ) \\ \\ x_1=(-25+171)/(2) \\ x_1=(146)/(2) \\ x_1=73 \\ \\ x_2=(-25-171)/(2) \\ x_2=(-196)/(2) \\ x_2=-98 \end{gathered}

There are two solutions for x, we will be discarding x = -98, as there is no negative dimensions.

With x = 73, we have the following


\begin{gathered} \text{width }=73 \\ \text{length }=73+25=98 \end{gathered}

Therefore, the dimensions are 73 feet for width, and 98 feet for length.

User Calisha
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