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Activity 5 - Quadratic Solution
This Activity and the related Essays on the History of Mathematics are on-line at MATHGYM (http://www.mathtrak.com.au/mic/)
I thought you might like to solve a couple of quadratics using the geometric method.
Solve x2 - 15x + 36 = 0
Solve x2 - 20x + 64 = 0
Question 1:
Question 2:
Answer Question 1
Solution of x2 - 15x + 36 = 0
We start by drawing AB of length equal to 15 ( = a).
Next we bisect AB at C and construct the perpendicular CO of length equal to 6 (= b).
Draw the arc centre O of length equal to 7.5 ( = a/2) to cut AB at D.
Using triangle OCD and Pythagoras' theorem CD is of length 4.5.
The solution of the equation is x = 12 (AD) and x = 3 (DB).
Answer Question 2
Solution of x2 - 20x + 64 = 0
We start by drawing AB of length equal to 20 ( = a).
Next we bisect AB at C and construct the perpendicular CO of length equal to 8 (= b).
Draw the arc centre O of length equal to 10 ( = a/2) to cut AB at D.
Using triangle OCD and Pythagoras' theorem CD is of length 6.
The solution of the equation is x = 16 (AD) and x = 4 (DB).