Question

the length of a rectangle is 6in.more than than its width .Its area is 50 sq.in Find the width of the rectangle....

Answer 1

L=6+W
A=LW
sub 6+W for L
A=(6+W)(W)
expand
oh, A=50
50=W²+6W
minus 50 both sides
0=W²+6W-50
we gots to use quadratic formula
for
aW²+bW+c=0
W=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
we got
a=1
b=6
c=-50


W=$\frac{-(6)+/- \sqrt{(6)^2-4(1)(-50)} }{2(1)}$
W=$\frac{-6+/- \sqrt{36+200} }{2}$
W=$\frac{-6+/- \sqrt{236} }{2}$
W=$\frac{-6+/- 2\sqrt{59} }{2}$
W=$-3+/- \sqrt{59}$

W=$-3+ \sqrt{59}$ or W=$-3- \sqrt{59}$
aprox
W=-10.6811 or W=4.68115
can't be negative width

width=4.68115 or -3√59