Answer:
The width is [tex](-5+\sqrt{77})\ yd[/tex]
Step-by-step explanation:
Let
x-----> the length of rectangle
y-----> the width of rectangle
we know that
The area of rectangle is equal to
[tex]A=xy[/tex]
[tex]A=52\ yd^{2}[/tex]
so
[tex]52=xy[/tex] ------> equation A
[tex]x=y+10[/tex] -----> equation B
substitute equation B in equation A
[tex]52=(y+10)y[/tex]
[tex]y^{2}+10y-52=0[/tex]
Solve the quadratic equation
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]y^{2}+10y-52=0[/tex]
so
[tex]a=1\\b=10\\c=-52[/tex]
substitute in the formula
[tex]y=\frac{-10(+/-)\sqrt{10^{2}-4(1)(-52)}} {2(1)}[/tex]
[tex]y=\frac{-10(+/-)\sqrt{308}} {2}[/tex]
[tex]y1=\frac{-10(+)2\sqrt{77}}{2}[/tex] -------> the solution is the positive value
[tex]y2=\frac{-10(-)2\sqrt{77}}{2}[/tex]
The width is [tex]\frac{-10(+)2\sqrt{77}}{2}\ yd=(-5+\sqrt{77})\ yd[/tex]
