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- Feb 14, 2012
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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 \(\displaystyle 19q+99p\).
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Evaluate 19q + 99p
- Thread starter anemone
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- #2
- Feb 7, 2012
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Re: Evaluate 19q+99p
$$\alpha\beta + \beta\gamma + \gamma\alpha = 9/2,\qquad \alpha\beta + (\alpha+\beta)(\gamma-8) = -7/2.$$ 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.
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.
- Jan 26, 2012
- 183
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.
I thought the same but my way was a little longer. There must be a clever way at getting to the answer.
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- Feb 14, 2012
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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 \(\displaystyle 19q+99p=19(15)+99(-9)=-606\)
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 \(\displaystyle 19q+99p=19(15)+99(-9)=-606\)
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