Answer :
The width of the rectangle = x.
The length of the rectangle is 3 yd more than twice the width = 2x+3.
The area is 35 yd².
The area is the width times the length.
[tex]35=x(2x+3) \\ 35=2x^2+3x \\ 0=2x^2+3x-35 \\ 0=2x^2+10x-7x-35 \\ 0=2x(x+5)-7(x+5) \\ 0=(2x-7)(x+5) \\ 2x-7=0 \ \lor \ x+5=0 \\ 2x=7 \ \lor \ x=-5 \\ x=\frac{7}{2} \ \lor \ x=-5[/tex]
The length and width must be positive so [tex]x=\frac{7}{2}=3.5[/tex].
[tex]2x+3=2 \times \frac{7}{2} + 3=7+3=10[/tex]
The dimension of the rectangle are 3.5 yd and 10 yd.
The length of the rectangle is 3 yd more than twice the width = 2x+3.
The area is 35 yd².
The area is the width times the length.
[tex]35=x(2x+3) \\ 35=2x^2+3x \\ 0=2x^2+3x-35 \\ 0=2x^2+10x-7x-35 \\ 0=2x(x+5)-7(x+5) \\ 0=(2x-7)(x+5) \\ 2x-7=0 \ \lor \ x+5=0 \\ 2x=7 \ \lor \ x=-5 \\ x=\frac{7}{2} \ \lor \ x=-5[/tex]
The length and width must be positive so [tex]x=\frac{7}{2}=3.5[/tex].
[tex]2x+3=2 \times \frac{7}{2} + 3=7+3=10[/tex]
The dimension of the rectangle are 3.5 yd and 10 yd.
