Coffin Problem

Algebra Level 3

2 2 y 1 3 = y 3 + 1 2\sqrt[3]{2y-1}=y^3+1

Solve the equation above for real y y . Assuming the roots are α \alpha , β \beta , and γ \gamma , submit α + β + γ \alpha +\beta +\gamma .


The answer is 0.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

On removing the radical, the equation reduces to :

y 9 + 3 y 6 + 3 y 3 16 y + 9 = 0 y^9+3y^6+3y^3-16y+9=0 .

The three real solutions of this equation are 1 , 5 1 2 , 5 1 2 1,\dfrac{\sqrt 5-1}{2},\dfrac{-\sqrt 5-1}{2} , and their sum is 1 2 × 1 2 = 0 1-2\times \dfrac{1}{2}=\boxed 0 .

Hi Mr. Alak, can you please elaborate on how you found the roots?

Mahdi Raza - 1 year ago

Log in to reply

2 y 1 3 = y 3 + 1 2 \displaystyle\sqrt[3]{2y-1}=\frac{y^3+1}{2}

(Note that the Equation is of the form f ( f ( x ) ) = x f(f(x))=x ,or alternatively f ( x ) = f 1 ( x ) f(x)=f^{-1}(x) , and since f ( x ) f(x) is an increasing function, f ( x ) = x f(x)=x , will provide solutions to the original equation, this can also be observed from the graph of f ( x ) = f 1 ( x ) f(x)=f^{-1}(x) ).

x 3 2 x + 1 = 0 x^3-2x+1=0 , x = 1 , 1 ± 5 2 \displaystyle x=1,\frac{-1\pm\sqrt{5}}{2}

Sarthak Sahoo - 1 year ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...