Evaluate 19q + 99p: 2x^3-8x^2+9x+p=0

[anemone]

Two of the roots of the equation $$2x^3-8x^2+9x+p=0$$ are also roots of the equation $$2x^3+8x^2-7x+q=0$$. Evaluate $$19q+99p$$.

 

[Opalg]

Re: Evaluate 19q+99p

Two of the roots of the equation $$2x^3-8x^2+9x+p=0$$ are also roots of the equation $$2x^3+8x^2-7x+q=0$$. Evaluate $$19q+99p$$.

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.

 

[Mathwebster]

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]

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 $$19q+99p=19(15)+99(-9)=-606$$

 

[MarkFL]

Re: Evaluate 19q+99p

I must applaud you for posting your solution...I know from experience that it can be daunting to post a solution when others have shown a shorter or more clever route. (Sun)