Re: Evaluate 19q+99p
Hmm...now that I see how Opalg approached the problem, I have to admit that this problem serves little purpose and is a weak problem.
Hi Jester, I think what Opalg has given here is the smartest and shortest solution but having said this, I will also show my solution in this post.
Let a, b, and m be the roots of the equation $$2x^3-8x^2+9x+p=0$$ and a, b and k be the roots of the equation $$2x^3+8x^2-7x+q=0$$.
We see that the sum of the roots for both equations are:
$$a+b+m=4$$
$$a+b+k=-4$$
Subtracting the second equation from the first equation, we obtain:
$$m-k=8$$
and the product of the roots for both equations are:
$$abm=-\frac{p}{2}$$, $$abk=-\frac{q}{2}$$
Dividing these two equations, we obtain:
$$\frac{m}{k}=\frac{p}{q}$$
By Newton Identities, we have:
$$(a^2+b^2+m^2)(2)+(-8)(4)+2(9)=0\implies a^2+b^2+m^2=7$$
$$(a^2+b^2+k^2)(2)+(8)(-4)+2(-7)=0\implies a^2+b^2+k^2=23$$
Subtracting these two equations yields:
$$k^2-m^2=16$$
$$(k+m)(k-m)=16$$
$$(k+m)(-8)=16$$
$$m+k=-2$$
Solving the equations $m+k=-2$ and $m-k=8$ for both $m$ and $k$, we get $m=3$ and $k=-5$.
When $m=3$, $$a+b+3=4\implies a+b=1$$
Substituting $m=3,\,k=-5,\,a+b=1$ back into the equations $$a^2+b^2+m^2=7$$, $$abm=-\frac{p}{2}$$ and $$abk=-\frac{q}{2}$$, we see that:
$$(a+b)^2-2ab+(3)^2=7$$
$$(1)^2-2ab+(3)^2=7$$
$$ab=\frac{3}{2}$$
Hence, $$\frac{3}{2}(3)=-\frac{p}{2}\implies p=-9$$ and $$\frac{3}{2}(-5)=-\frac{q}{2}\implies q=15$$.
And this gives $$19q+99p=19(15)+99(-9)=-606$$
