Answer to one mathleague quesiton from AoPS

Question is here.

You have two congruent triangles. 17-same angle and- x (SAS)
Using distance formula, the two green lines are of the same length.
\(\left( a-25\right) ^{2}+\left( 20-15\right) ^{2}=a^{2}+20^{2}\)
a = 5

Use another distance formula to get x -- the blue line.
\(\sqrt {\left( 5-17\right) ^{2}+20^{2}}=\sqrt {544}= 4\sqrt {34}\)

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.