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.

Find the area of the petal, or the football shape.

Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.

Circle and area revisited from Mathcounts mini

The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.

Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)

The overlapping part is C.

A + B - C = 6^2 so C = A + B - 36 or 18pi - 36





                                                                                               

Solution IV:

If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.

Use the area of the square minus that will again give you the answer.
$6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)$
= 18pi - 36




Similar triangles and triangles that share the same vertexes/or/and trapezoid

Another link from my blog

Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.

Take care and happy problem solving !!