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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?

Respuesta :

luisejr77

Answer: 5 inches.

Step-by-step explanation:

The  formula for calculate the area of a rectangle is:

[tex]A=w*l[/tex]

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:

[tex]l=2w+3[/tex]

Substitute [tex]l=2w+3[/tex] and the area into the formula and solve for the width:

[tex]65=(2w+3)w\\65=2w^2+3w\\0=2w^2+3w-65[/tex]

Use the quadratic formula:

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

Choose the positive value.

Therefore, the width is: 5 inches.

kudzordzifrancis

Answer:

The width is 5in.

Step-by-step explanation:

Let the width of the rectangle be [tex]w\:in.[/tex].

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

[tex]l=(2w+3)in[/tex]

The area of the rectangle is given by;

[tex]Area=lw[/tex]

it was given that the area is 65 sq in.

This implies that;

[tex]65=(2w+3)w[/tex]

Expand the brackets

[tex]65=2w^2+3w[/tex]

[tex]2w^2+3w-65=0[/tex]

Factor

[tex](2w+13)(w-5)=0[/tex]

[tex](2w+13)=0,(w-5)=0[/tex]

[tex]w=-\frac{13}{2},w=5[/tex]

Discard the negative value.

Hence [tex]w=5in.[/tex]

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