PYTHAGOREAN QUADRATIC

Pythagorean Quadratic



Pythagorean Quadratic

Solution to Question 98 (Pg # 371)

Using the Pythagorean Theorem

H2 = P2 + B2

According to the given condition

(2x + 6)2 = (2x + 4)2 + x2

4x2 + 24x + 36 = 4x2 + 16x + x2

x2 - 8x - 20 = 0

Using the zero factor property

x2 - 10x + 2x - 20 = 0

x (x - 10) + 2 (x - 10) = 0

(x + 2) (x - 10) = 0

Either (x + 2) = 0or (x - 10) = 0

x = -2and x = 10

Therefore, the value x = 10 shows the real solution to ...