Difference between revisions of "2014 AIME I Problems/Problem 9"

Revision as of 18:55, 14 March 2014

Problem 9

Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ .

Solution

Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$ . Find $x_2(x_1+x_3)$ .

We note that $\dfrac{1}{\sqrt{2014}}$ is a solution since $(\dfrac{1}{\sqrt{2014}})^3*\sqrt{2014}-4029(\dfrac{1}{\sqrt{2014}})^2+2=\dfrac{1}{2014}-\dfrac{4029}{2014}+2=\dfrac{1-4029+4028}{2014} = 0$

We claim that $x_2=\dfrac{1}{\sqrt{2014}}$ [please clarify]

By Vieta's formula we have that the $x^2$ coefficent is equal to $-x_1-x_2-x_3$ and that the $x$ coefficent is equal to $x_1x_2+x_1x_3+x_2x_3$ . Using the values in the above equation we get: $-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029$

@@ Line 10: / Line 10: @@
 We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math>
+[please clarify]
-by Vieta's formula we have that the <math>x^2</math> coefficent is equal to <math>-x_1-x_2-x_3</math> and that the <math>x</math> coeefficent is equal to <math>x_1x_2+x_1x_3+x_2x_3</math> so using the values in the above equation we get:
+By Vieta's formula we have that the <math>x^2</math> coefficent is equal to <math>-x_1-x_2-x_3</math> and that the <math>x</math> coefficent is equal to <math>x_1x_2+x_1x_3+x_2x_3</math>. Using the values in the above equation we get:
 <math>-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029</math>

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Difference between revisions of "2014 AIME I Problems/Problem 9"

Revision as of 18:55, 14 March 2014

Problem 9

Solution

See also