The length of a rectangle is 2m more than the width. The area of the rectangle is 20 m^2. Find the dimensions of the rectangle, to the nearest tenth of a metre.
The answer is 3.6 m by 5.6m... how?
We know that for a rectangle, the area is length x width A = l*w We know that the length is 2 meters more than the width So the the length minus 2 in the width => l - 2 = w l = w + 2 Substitutin in the area formula: A = (w+2)(w) 20 = w(w+2) Solve for w using the quadratic equation, then using the w you can find l
solve for w using the quadratic equation? so it's going ot be w^2 + 2w - 20 = 0
yup
ohhhh,
Get it now? :)
yup! Oh, but what's a product of -20 and a sum -20? o.o
Product of -20, sum of 2... Do you know how to use the quadratic equation?
sort of.. isn't it like factoring?
Its like factoring, but when like integer numbers dont work to solve the equation http://www.oncalc.com/wp-content/uploads/2011/02/quadratic-equation-calculator.gif
You see that equation before?
ohhhh the quadratic formula?
Yeah, sorry didnt know what you call it
it's okay. But what about it? That's all I use after I've figured it's w^2 + 2w - 20 =0?
Yup, plug it in to the quadratic formula
ohh, thank you!
Okay, well I got -2 +/- 2sqr.rt:21/2 ...
punch that into a calculator and what do you get?
so I don't have to do 2(-2 +/- sqr.rt:21/2?
well what number do you get from that?
Like evaluate the expression
not the answer.
(-2 + sqrt ( 4 - 4(1)(-20) )/ 2
I got 3.58
if you do (-2 - sqrt ( 4 - 4(1)(-20) )/ 2 , you get a negative number, and can omit it because there is no such thing as negaive lengths
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