Linear Algebra problems im having

joey12345 · 17-05-2011 07:52PM

ok so i got the determinant and ended up with

x^3-25x+66=0

i no that x+6 is a factor and then x=-6 so i get -66+66=0

should i do any more checks or would that be all do you think?

Ficheall · 17-05-2011 07:57PM

What you have shows that -6 is a real root, but you must also show that the other roots are imaginary.

If you divide x^3-25x+66 by x+6 you should be left with a quadratic with imaginary roots. Only then can you say that x+6 is the only real root.

joey12345 · 17-05-2011 07:58PM

ahh yes thank you ill do that now

joey12345 · 17-05-2011 08:25PM

ah yea ok so when i divided in i got
x^2-6x+11

so because this cant be done off the top of your head like more easy quadratics that would mean ive to use the quadratic formula,get them in the form a+bi so -6 IS the only real root?

Michael Collins · 17-05-2011 08:28PM

joey12345 wrote: »

ah yea ok so when i divided in i got
x^2-6x+11

so because this cant be done off the top of your head like more easy quadratics that would mean ive to use the quadratic formula,get them in the form a+bi so -6 IS the only real root?

Correct. To be extra precise you could show that the quadratic discriminant is negative: [latex] b^2 - 4ac < 0 [/latex] which means there can be no real roots.

joey12345 · 17-05-2011 08:34PM

thanks man. the exams tomorrow and Q1 is built up of easy enough Q's so just getting the seperate questions out so ive a better idea and then hopefully ill be fine.

Linear Algebra problems im having

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