^{2}+50x = 2400

^{2}+50x-2400 = 0

**Hence the width of the field is 30m**

A retangular field covers 2400m^2. its lengh is 50m longer than it's width what's its with

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Mohammed R. | Specialist in GED and College Prep MathSpecialist in GED and College Prep Math

Marked as Best Answer

Let the width of the field = x

So, the length of the field = x+50

We know,

area of rectangle = length*width

or 2400 = x*(x+50)

x^{2}+50x = 2400

x^{2}+50x-2400 = 0

(x+80) (x-30) = 0

So, x = -80 or x = 30

Since we can't have negative width, the answer is 30

Shouldn't the second line below have -2400 not + 2400?

x2+50x = 2400

x2+50x+2400 = 0

x2+50x+2400 = 0

Second line should be

x^{2} + 50x - 2400 = 0

Nevertheless, you got the correct solution. But you should go back to fix that so the student and any other person reading your solution does not get thrown off or confused with your process.

Thank you because I got that solution but I didn't get the same processes

Thank you Mark and Michael for pointing out the error. I've just fixed it.

Hope the solution helped you Haylen

You are given the area of 2400m^{2}. The area of a rectangle is length times width.

Let width = x

Let length = x + 50

Using these variables, we can create an equation that represents the area.

x(x + 50) = 2400

Solve for x.

Distribute the x to get rid of parenthesis.

x^{2} + 50x = 2400

Subtract 2400 on both sides of the equation.

x^{2} + 50x - 2400 = 0

Solve this quadratic equation for x. You may need to use the quadratic formula if you cannot easily find two integers that multiply to -2400 and add to 50.

Then, select the positive value of x.

Another way is to remove the zero digits of the integers in the equation to get

x^{2} + 5x - 24 = 0

This makes it easier to solve. We know that 8 and -3 are two integers that add up to 5 and multiply to -24. So if we add one zero to each integer, we get 80 and -30 respectively.

80 - 30 = 50 ----> middle coefficient

(80)(-30) = -2400 ----> last term

Factored form is then

(x + 80)(x - 30) = 0

Whenever you have terms that are multiples of 10, try this technique to avoid taking too much time factoring out very large numbers.

This is a nice trick Michael. I didn't know this. Just learnt something nice. Thank you for sharing.

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