Showing posts with label sum and product of roots.. Show all posts
Showing posts with label sum and product of roots.. Show all posts

Thursday, December 5, 2013

Sum and Product of roots : Vieta -- > Questions from 2010-2011 Mathcounts Super Stretch

Questions: (detailed solutions below)

#1 : What is the sum of the solutions of 6x2 + 5x - 4 = 0 ? Express your answer as a common fraction. 

#2 : A quadratic equation of the form x2 + kx + m = 0 has solutions x = 3 + 2 2  and 3 - 2 2 
What is the value of k + m? 

#3 : What is the sum of the reciprocals of the solutions of 4x2 - 13x + 3 = 0 ? Express your answer as a common fraction. 













Solutions : 

#1:  6x2 + 5x -4 = 0 divided the whole equation by 6 and you have x2 + (5/6) x - 4/6 = 0, which means that the sum of the solutions is - 5/6. 

#2: The two roots are 3 + 2 2  and 3 - 2 2 , which means that -k = 3 + 2 2 + 3 - 2 2 
k = -6;  m = (3 + 2 2 ) (3 - 2 2 ) = 9 - 8 = 1 so m + k = -6 + 1 = -5

#3:
Solution I:
4x2 - 13x + 3 = 0; divided the whole equation by 4 and you have  x2 - (13/4)x + 3/4 = 0,
which means that the sum of the two roots, if they are x and y, are 13/4 and their product is 3/4.

1/x + 1/y = (x + y) / xy = 13/4  divided by 3/4 = 13/3

Solution II: Tom shows Rob and Rob shows me how to solve this using another method.

The original equation is 4x2 -13x + 3 = 0 To find the sum and product of the reciprocals, you flip the equation so it becomes 3x2 - 13x + 4 = 0

Using the same way you find the sum of the two roots,the answer is 13/3.