Question

2. The length of a rectangular floor is 5 feet longer than its width w. The area of the floor is 535 ft2.

Answer 1

Answer:

Solution given:

let width of the floor be w.

length of floor =5ft+w

Area of floor=535ft²

we have

Area of floor =length*width

substituting value

535=(5+w)*w

opening bracket

535=5w+w²

w²+5w-535=0

Comparing above equation with ax²+bx+c=0,we get

a=1

b=5

c=-535

In terms of W in quadratic equation is:

W=$\frac{-b±\sqrt{b²-4ac}}{2a}$

Substituting value

w=$\frac{-5±\sqrt{5²-4*1*-535}}{2*1}$

Solving like terms

$\frac{-5±\sqrt{2165}}{2}$

=Taking positive

w= $\frac{-5+\sqrt{2165}}{2}$

w=20.76

taking negative

-25.76(neglected)

So

width of floor=20.76ft

again

length =20.76+5=25.76ft

quadratic equation in terms of w that represents the situation is w= $\frac{-5+\sqrt{2165}}{2}$

The dimension of the floor is:

length =20.76+5=25.76ft

width of floor=20.76ft

Answer 2

Let width be x

• Length=x+5

$\\ \rm\Rrightarrow Area=Length(Breadth)$

$\\ \rm\Rrightarrow x(x+5)=535$

$\\ \rm\Rrightarrow x^2+5x=535$

$\\ \rm\Rrightarrow x^2+5x-535=0$

Solving further we get

$\\ \rm\Rrightarrow x=\pm 20.7$

$\\ \rm\Rrightarrow Breadth=20.7ft$

$\\ \rm\Rrightarrow Length=20.7+5=25.7$