suppose two sides of a square are given as x + 3 and 12 - x. What is its area ?
Area=(x+3)*(12-x) =-x^2+9x36
solve= -x^2+9x+36
Well, all sides of a square are equal in length, so no exception for these 2, they're equal as well :P x+3 = 12 - x Solving this, you can find out 'x', then substitute it in the equation for any side, i.e. (x+3) or (12-x) and find out the length of the sides. Squaring the side-length gives you the area of the square, as you may know. Solve and enjoy!
\[\huge \color{green}{Area = Side^2}\]
You have to find x from here as @apoorvk told you.. Area will be in constant form and not in variable form..
@burakb am not sure equating that quadratic to zero is correct. It just represents the area, and 'x' has just one possible value actually.
If you equated the quadratic to 0 and solved it you'd be solving for the point in which the area of the square was 0.
But this question doesn't have any restriction.So i think so just we should find that quadratic equation
@Malaria exactly what I meant! @burakb think about this!
sorry i didnt see that is the square So, x+3=12-x Then 2x=9 İf only if x=(2/9) so area must be (9/2)*(9/2)=81/4
x=(9/2)
We can use just x=9/2 equation ...
@FlyinSolo_424 do you realise that factoring that quadratic led you BACK to the same factors which you had just 'distributed'?
Yeah we just put in x=9/2 in x+3, or 12-x (either will do since both are equal) and find out the side length. So, side length = 9/2 + 3 = 15/2 = 7.5 So, area = \((7.5)^2\) sq. units.
My bad dude. I didnt realize. But anyways, Ive been half awake and have gotten no sleep and been doing homework alll night.. give me a little slack. Its not like I gave her a wrong answer then if I ended back to where I was.
@apoorvk
Hey it's completely alright, we all make mistakes ;) (good that you tried)
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