# Evaluate 19q + 99p

#### anemone

##### MHB POTW Director
Staff member
Two of the roots of the equation $$\displaystyle 2x^3-8x^2+9x+p=0$$ are also roots of the equation $$\displaystyle 2x^3+8x^2-7x+q=0$$. Evaluate $$\displaystyle 19q+99p$$.

#### Opalg

##### MHB Oldtimer
Staff member
Re: Evaluate 19q+99p

Two of the roots of the equation $$\displaystyle 2x^3-8x^2+9x+p=0$$ are also roots of the equation $$\displaystyle 2x^3+8x^2-7x+q=0$$. Evaluate $$\displaystyle 19q+99p$$.
Use Vieta's relations. The sum of the roots of the first equation is $4$, and the sum of the roots of the second equation is $-4$. So if the roots of the first equation are $\alpha,\ \beta$ and $\gamma$, then the roots of the second equation are $\alpha,\ \beta$ and $\gamma-8$. Vieta's relations tell us that
$$\alpha\beta + \beta\gamma + \gamma\alpha = 9/2,\qquad \alpha\beta + (\alpha+\beta)(\gamma-8) = -7/2.$$ Therefore $-\frac72 = \frac92 - 8(\alpha+\beta)$, from which $\alpha+\beta = 1$. But $\alpha + \beta + \gamma = 4$, and so $\gamma=3$ (and $\gamma-8 = -5$).

Putting $x=3$ as a solution to the first equation, you find that $p=-9$; and putting $x=-5$ as a solution to the second equation, you find that $q=15$.

You can then find $19q + 99p = -606$ but that seems a bit pointless.

#### Jester

##### Well-known member
MHB Math Helper
Re: Evaluate 19q+99p

I thought the same but my way was a little longer. There must be a clever way at getting to the answer.

#### anemone

##### MHB POTW Director
Staff member
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 $$\displaystyle 2x^3-8x^2+9x+p=0$$ and a, b and k be the roots of the equation $$\displaystyle 2x^3+8x^2-7x+q=0$$.

We see that the sum of the roots for both equations are:

$$\displaystyle a+b+m=4$$

$$\displaystyle a+b+k=-4$$

Subtracting the second equation from the first equation, we obtain:

$$\displaystyle m-k=8$$

and the product of the roots for both equations are:

$$\displaystyle abm=-\frac{p}{2}$$, $$\displaystyle abk=-\frac{q}{2}$$

Dividing these two equations, we obtain:

$$\displaystyle \frac{m}{k}=\frac{p}{q}$$

By Newton Identities, we have:

$$\displaystyle (a^2+b^2+m^2)(2)+(-8)(4)+2(9)=0\implies a^2+b^2+m^2=7$$

$$\displaystyle (a^2+b^2+k^2)(2)+(8)(-4)+2(-7)=0\implies a^2+b^2+k^2=23$$

Subtracting these two equations yields:

$$\displaystyle k^2-m^2=16$$

$$\displaystyle (k+m)(k-m)=16$$

$$\displaystyle (k+m)(-8)=16$$

$$\displaystyle m+k=-2$$

Solving the equations $m+k=-2$ and $m-k=8$ for both $m$ and $k$, we get $m=3$ and $k=-5$.

When $m=3$, $$\displaystyle a+b+3=4\implies a+b=1$$

Substituting $m=3,\,k=-5,\,a+b=1$ back into the equations $$\displaystyle a^2+b^2+m^2=7$$, $$\displaystyle abm=-\frac{p}{2}$$ and $$\displaystyle abk=-\frac{q}{2}$$, we see that:

$$\displaystyle (a+b)^2-2ab+(3)^2=7$$

$$\displaystyle (1)^2-2ab+(3)^2=7$$

$$\displaystyle ab=\frac{3}{2}$$

Hence, $$\displaystyle \frac{3}{2}(3)=-\frac{p}{2}\implies p=-9$$ and $$\displaystyle \frac{3}{2}(-5)=-\frac{q}{2}\implies q=15$$.

And this gives $$\displaystyle 19q+99p=19(15)+99(-9)=-606$$