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June 29th, 2018, 02:23 PM   #11
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Quote:
 Originally Posted by MathsLearner123 This is the attempt I have done. The final equation is $\displaystyle x(\lambda^3+4\lambda-40)=0;$ I am not sure why I got a different equation last time. Sorry if the attachments are not clear. Next time, I will try to use math editor. Please advise whether I am correct.
get rid of the factor of $x$ in front and you'll have it correct.

There is one real solution.

Last edited by skipjack; June 29th, 2018 at 09:32 PM.

June 29th, 2018, 09:51 PM   #12
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Quote:
 Originally Posted by MathsLearner123 The final equation is $\displaystyle x(\lambda^3+4\lambda-40)=0$ Please advise whether I am correct.
That is the correct equation, but $y$ and $z$ can be eliminated without dividing by anything that might be zero.

Note that $\displaystyle \lambda^3 + 4\lambda - 40$ is the determinant of the matrix of coefficients of $x$, $y$ and $z$ in the original three equations.

Can you prove that $\displaystyle \lambda^3 + 4\lambda - 40 = 0$ has only one real solution?

June 30th, 2018, 08:16 PM   #13
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Quote:
 Originally Posted by skipjack Can you prove that $\displaystyle \lambda^3 + 4\lambda - 40 = 0$ has only one real solution?
I tried from this https://en.wikipedia.org/wiki/Cubic_function
and calculated the discriminant which is less than 0 and hence it has one real root.
Discriminant is
$\displaystyle ax^3+bx^2+cx+d=0; \text{ discriminant } = 18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2; a=1, b=0, c=4,d=-40 =0 - 0 + 0 - 256 - 43200 =-43456$
Attached Images
 discriminant.jpg (20.2 KB, 2 views)

Last edited by skipjack; June 30th, 2018 at 11:01 PM.

 June 30th, 2018, 10:12 PM #14 Senior Member     Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044 Perhaps a simpler way of showing it has only one real root is to realize that $\lambda^3+4\lambda -40$ is strictly increasing.

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