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June 29th, 2018, 02:23 PM  #11  
Senior Member Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044  Quote:
There is one real solution. Last edited by skipjack; June 29th, 2018 at 09:32 PM.  
June 29th, 2018, 09:51 PM  #12  
Global Moderator Joined: Dec 2006 Posts: 19,191 Thanks: 1649  Quote:
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  
Member Joined: Aug 2017 From: India Posts: 42 Thanks: 1  Quote:
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 } = 18abcd4b^3d+b^2c^24ac^327a^2d^2; a=1, b=0, c=4,d=40 =0  0 + 0  256  43200 =43456 $ 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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