Question

The length of a rectangle is three feet shorter than twice it's width. If the area of the rectangle is 170ft^(2) , What is the length of the rectangle?

Thx.

Answer 1

Let the length of rectangle be \(l\) and width be \(w\) , it's given that length of rectangle is 3 feet shorter than twice the width
so
\[l=2w-3\]
Area is given as 170 ft^2 so
\[l w= 170\]
we have l= 2w-3, so
\[(2w-3)(w)=170\]
\[2w^2-3w=170\]
we have now
\[2w^2-3w-170=0\]
now we know that solution of a quadratic equation \[ax^2+bx+c=0\] is given by
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
now we have
\[2w^2-3w-170=0\]
here a=2 b=-3 and c=-170 so we get
\[w=\frac{-(-3)\pm \sqrt{(-3)^2-4\times 2 \times -170}}{2\times 2}\]
we get
\[w=\frac{3\pm \sqrt{9+1360}}{4}\]
we have now
[\[w=\frac{3\pm \sqrt{1369}}{4}\]
we know that \(\sqrt{1369}=37\)
so we get
\[w=\frac{3\pm 37}{4}\]
either
\[w=\frac{3+ 37}{4}or \frac{3- 37}{4}\]
since width can't be negative, we'llhave
\[w= \frac{3+37}{4}\]
we get
\[w=10\]
now we know
\[lw=170\]
w=10
so
\[l\times 10=170=>l=17\]
so we have l=17 feet and w=10 feet

Answer 2

@leidyr check this out, do you understand this?

Answer 3

Yes, well explained it. Thanks so much..