Christmas Questions: No.1

Algebra Level pending

Let $x_{1},x_{2},x_{3},....,x_{n}$ be the real solutions of the polynomial, $2x^4+9x^3+4x^2+9x+2=0$ such that $x_{1}< x_{2}< x_{3}< .... <x_{n}$ If $x_{n}-x_{1}$ can be expressed as $\frac{\sqrt{A}}{B}$ , with $A$ and $B$ being square-free positive integers. Calculate $A+B$

The answer is 67.

1 solution

Zach Bian
Jan 3, 2018

Using Geogebra, we can find the roots of the function: Root(2x^4+9x^3+4x^2+9x+2) .

There are only two roots: A(-4.27,0) and B(-.23,0), not to be confused with the problem's $A$ and $B$ .

Next input (x(B)-x(A))^2 , which gives a=16.25. This is equivalent to the problem's $\frac{A}{B^2}$ . In fractional form this is $\frac{65}{4}$ . This tells us that the problem's $A$ and $B$ are 65 and 2, respectively. The sum of which is 67.

Cool solution but could you solve it the algebraic way?

Kok Hao - 3 years, 5 months ago

Christmas Questions: No.1

The answer is 67.

1 solution

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