Linear Algebra problems im having

joey12345

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

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

ahh yes thank you ill do that now

joey12345

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

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

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.