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Let r, s, and t be the roots of the equation x^3 - 2x + 1 = 0 in some order. What is the maximal value of r^3 - s- t?

 Aug 11, 2019
 #1
avatar
+1

Solve for x:
x^3 - 2 x + 1 = 0

The left hand side factors into a product with two terms:
(x - 1) (x^2 + x - 1) = 0

Split into two equations:
x - 1 = 0 or x^2 + x - 1 = 0

Add 1 to both sides:
x = 1 or x^2 + x - 1 = 0

Add 1 to both sides:
x = 1 or x^2 + x = 1

Add 1/4 to both sides:
x = 1 or x^2 + x + 1/4 = 5/4
Write the left hand side as a square:
x = 1 or (x + 1/2)^2 = 5/4

Take the square root of both sides:
x = 1 or x + 1/2 = sqrt(5)/2 or x + 1/2 = -sqrt(5)/2

Subtract 1/2 from both sides:
x = 1 or x = sqrt(5)/2 - 1/2 or x + 1/2 = -sqrt(5)/2

Subtract 1/2 from both sides:


x = 1 or x = sqrt(5)/2 - 1/2 or x = -1/2 - sqrt(5)/2
1^3  - [sqrt(5)/2 - 1/2] - [ -1/2 - sqrt(5)/2] = 2

 Aug 11, 2019
edited by Guest  Aug 11, 2019
 #2
avatar+561 
+2

By observation, we can see that x = 1 is a root because (1)^3 - 2(1) + 1 = 1 - 2 + 1 = 0

If x = 1, then (x - 1) = 0

Assuming x^3 - 2x + 1 = 0 can be written as (x+a)(x+b)(x+c) = 0, where -a, -b, and -c are the roots, then we can divide the equation by the root we already know to simplify it.

(x^3 - 2x + 1) / (x - 1) = x^2 + x - 1

Use the quadratic equation to find the roots of x^2 + x - 1

x = (-b +/- (b^2 - 4ac)^(0.5)) / (2a) = (-(1) +/- ((1)^2 - 4(1)(-1))^(0.5)) / (2(1)) = -0.5 +/- (5^0.5) / 2

The other 2 roots are x = -0.5 + (5^0.5) / 2 ~= 0.6 and x = -0.5 - (5^0.5) / 2 ~= -1.6

The largest root is x = 1, so let r = 1, s = 0.6, t = -1.6

r^3 - s - t = (1)^3 - (0.6) - (-1.6) = 2

 

2 is the maximal value!

 Aug 11, 2019
 #3
avatar+106885 
0

Thanks to both of you.

 

It is good to see you again Will  laugh

Melody  Aug 12, 2019
edited by Melody  Aug 12, 2019

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