Question

The length of a rectangle is 3 in more than twice it’s width, and it’s area is 65 sq in. What is it’s width?

Answer 1

Answer:

The width is 5in.

Step-by-step explanation:

Let the width of the rectangle be $w\:in.$.

The length of a rectangle is 3 in more than twice it’s width, so we write the equation;

$l=(2w+3)in$

The area of the rectangle is given by;

$Area=lw$

it was given that the area is 65 sq in.

This implies that;

$65=(2w+3)w$

Expand the brackets

$65=2w^2+3w$

$2w^2+3w-65=0$

Factor

$(2w+13)(w-5)=0$

$(2w+13)=0,(w-5)=0$

$w=-\frac{13}{2},w=5$

Discard the negative value.

Hence $w=5in.$

Answer 2

Answer: 5 inches.

Step-by-step explanation:

The  formula for calculate the area of a rectangle is:

$A=w*l$

Where l is the length and w is the width.

If  the length of a rectangle is 3 inches more than twice it’s width, then:

$l=2w+3$

Substitute $l=2w+3$ and the area into the formula and solve for the width:

$65=(2w+3)w\\65=2w^2+3w\\0=2w^2+3w-65$

Use the quadratic formula:

$w=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\w=\frac{-3\±\sqrt{3^2-4(2)(-65)}}{2(2)}\\\\w=5\\w=-6.5$

Choose the positive value.

Therefore, the width is: 5 inches.