Super hard problem - answer needed (1 Viewer)

[Valsimot]

Ive got a maths assignment and i cant do the last question its driving me nuts, the question is

(a^2+b^2)x^2 + 2b(a+c)x + (b^2+c^2) = 0, Find the relationship between a, b and c

Could someone please post a solution and explain to me thanks a lot.

 

[vafa]

(a^2+b^2)x^2+2b(a+c)x+(b^2+c^2)=0

Delta(discriminant)= 4b^2(a+c)^2-4(a^2+b^2)(b^2+c^2)

Delta=8acb^2-4a^2c^2-4b^4

Delta= -4((b^2-ac)^2)

Delta=<0

b^2-ac=>0 or b^2-ac=<0

b^2=>ac or
b^2=<ac

 

[Valsimot]

Thanks heaps, but could you tell me, does it mean only b^2=ac, or b^2=>ac and b^2<=ac?

 

[vafa]

b^2=>ac or b^2=<ac

 

[vafa]

Thanks heaps, but could you tell me, does it mean only b^2=ac, or b^2=>ac and b^2<=ac?

keep in mind onething and that is the or which we use in English is quite different to the or which we use in mathematics.

you need to have or.

b^2=>ac or b^2=<ac

i am writing two iequalitis one is bigger than or equal to and the other one is less than or equal to and i guess that the other one is not going to publish. just note that

 

[Valsimot]

Thanks heaps, i woulnd't have thought of using the discriminant. The answer just confuses me because you can have both b^2=>ac or b^2<=ac, which is sort of stupid because you can then just pick any value for a b and c.

 

[vafa]

you have (b^2-ac)^2=>0 and so b^2-ac=>0 or b^2-ac=<0 because we do not know that b^2-ac is bigger than or equal to 0 or it is less than or equal to zero. We may have both cases...!

 

[vafa]

but you can not say that b and c can not be complex, they can be anything and in that case discriminant is less than zero and i guess that you are not allowed to make any assumptions since the coefficients are in general form and they can be anything and if you consider them in general form you will get the same result as i got
ok lets say a=b=c=0, then obviously that would be a really funny question...!

 

[vafa]

I agree...!