2015 Countdown Problem 20: A Cubic Expansion in a Quadratic Equation

This is the solution for the problem without solving individually for $\alpha$ and $\beta$ .

Since $\alpha$ and $\beta$ are both roots of the quadratic equation, and the leading coefficient is one, the equation can be written in the form

$(x-\alpha )(x-\beta )=0={ x }^{ 2 }-35x+300$

Expanding, we get

${ x }^{ 2 }-\alpha x-\beta x+\alpha \beta ={ x }^{ 2 }-35x+300\\{ x }^{ 2 }-(\alpha +\beta )x+\alpha \beta ={ x }^{ 2 }-35x+300$

Therefore,

$\alpha +\beta =35\\ \alpha \beta =300$

Taking the first equation and squaring both sides, we get

${ \left( \alpha +\beta \right) }^{ 2 }={ 35 }^{ 2 }$

Which expands to

${ \alpha }^{ 2 }+2\alpha \beta +{ \beta }^{ 2 }=1225$

Substituting 300 in for $\alpha \beta$ , we get

$\alpha^{ 2 }+2\left( 300 \right) +\beta ^{ 2 }=1225\\\alpha ^{ 2 }+600+\beta ^{ 2 }=1225\\\alpha ^{ 2 }+\beta ^{ 2 }=625$

Now, let us look at the second quadratic equation, ${ x }^{ 2 }-mx+n=0$ . There is only one real root, which is $x={ \alpha }^{ 3 }+{ \beta }^{ 3 }$ .

This can be factored to get

$x=\left( { \alpha }+{ \beta } \right) \left( { \alpha }^{ 2 }-\alpha \beta +{ \beta }^{ 2 } \right) \\ x=\left( { \alpha }+{ \beta } \right) \left( \left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) -\alpha \beta \right)$

Substituting the values that we solved for earlier, we get

$x=\left( 35 \right) \left( 625-300 \right)$

Which gives us

$x=11375={ \alpha }^{ 3 }+{ \beta }^{ 3 }$

Now, since $x=11375$ is the only root of this quadratic equation, it can be written in the form of

${ \left( x-11375 \right) }^{ 2 }=0={ x }^{ 2 }-mx+n$

Expanding, we get

${ x }^{ 2 }-22750x+129390625={ x }^{ 2 }-mx+n$

Therefore, $n=129390625$

To get the final solution to the problem, we need to add up all of the digits in n. This is done as such:

$1+2+9+3+9+0+6+2+5=\boxed{37}$

By Vieta's Formulas:

$\alpha+\beta=35$

$\alpha \beta=300$

Now, as $x^2-mx+n$ only has one real root can be factoring as perfect square $\therefore x^2-mx+n=(x-(\alpha^3+\beta^3))^2=x^2-2(\alpha^3+\beta^3)x+n$

$\alpha^3+ \beta^3=(\alpha+ \beta)^3-3(\alpha+ \beta)(\alpha \beta)=35^3-3(35)(300)$

As $x^2-mx+n$ is a perfect square binomial its discriminant is $0$

$m=2(11375)$

$m^2-4(1)(n)=0$

$4(11375)^2=4n$

$n=11375^2=129390625$

$1+2+9+3+9+6+2+5=\boxed{37}$