The length of a rectangle is 10 more than it’s width. If the area of the rectangle is 52yd^2 , find the appropriate width.

Question

asked Sep 22, 2020 193k views

2 Answers

Carlos Landeras · Answer 1 · 2020-09-29T10:20:42+0000

Answer:

The width is
$(-5+√(77))\ yd$

Explanation:

Let

x-----> the length of rectangle

y-----> the width of rectangle

we know that

The area of rectangle is equal to

A=xy

$A=52\ yd^(2)$

so

52=xy ------> equation A

x=y+10 -----> equation B

substitute equation B in equation A

52=(y+10)y

y^(2)+10y-52=0

Solve the quadratic equation

The formula to solve a quadratic equation of the form
ax^(2) +bx+c=0 is equal to

$x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}$

in this problem we have

y^(2)+10y-52=0

so

$a=1\\b=10\\c=-52$

substitute in the formula

$y=\frac{-10(+/-)\sqrt{10^(2)-4(1)(-52)}} {2(1)}$

$y=\frac{-10(+/-)√(308)} {2}$

y1=(-10(+)2√(77))/(2) -------> the solution is the positive value

y2=(-10(-)2√(77))/(2)

The width is
$(-10(+)2√(77))/(2)\ yd=(-5+√(77))\ yd$