Masterscorp wrote:
gvij2017 wrote:
l = 3s + 2
Area of triangle: ls/2
400 = s*(3s+2)/2
3s^2 + 2s = 800
s=16 is answer to this equation.
Is there any convenient way to solve the equation \(3s^2 + 2s = 800\) ? I transformed it to \(s² + \frac{2b}{3}- \frac{800}{3}\) and used the pq-method. The problem is that I had to find \(\sqrt{2401}\), which equals 49. However, determining that is it 49 is not that simple.
Thats why I am asking whether there is an easier way. Thanks for any help!
Masterscorp , the \(pq\) method is not easy here. If you can use the pq method, the two suggestions below will be easy.
To find the square root of any difficult perfect square, you can
1) think about
easy perfect squares that are "close" to 2401:
\(2500 = 50^2\)
\(1600 = 40^2\)
The square root of 2401 is between 40 and 50
2500 and its square root, 50, are too great, but not by much. Try 49.
\(49^2=2401\) OR: the units digit of \(\sqrt{2401}\) is either 1 or 9
Try 49 (i.e., instead of 41, because \(41^2\) will be too close to 1600)
2) Find prime factors of 2401
2401 is not divisible by 2, 3, or 5. Try 7:
\(\frac{2401}{7}=343\)
343 is not divisible by 2, 3, or 5. Try 7:
\(\frac{343}{7}=49=(7*7)\)
We have 2401 = (7*7*7*7) = (49*49), that is
\(\sqrt{2401}=\sqrt{49*49}=49\)
Done
Wanting to understand is smart.
On other occasions, JMO, it's better to use a mix of understanding and available shortcuts - in this case, some equation (not necessarily a quadratic, which I avoided) + answer choices.
Hope that helps.