Question

Prove the following statement.

$\large \textup{Let a and b be nonzero integers. If } a+b\neq 3 \textup{ and } 9a+b\neq 1, \textup{then the equation } ax^3+bx-3=0 \textup{ does not have a solution that is a real number.}$

Answer 1

I with real Let f(x) = ax'+bc-3, a, b ea. Then f(x) is a cold odd degree polynomel wit Coefficient Then for a large value of x bay x=M, M60) EIR & for ayo we have f(M)= am² +6M-370. & f(-M)= - am² + b M-3 Lo Then by Intermediate value theorem 7 CERC-MM) s.t. f(c) = 0. c'is a real noot of f(x) = 0. Similarly for a co there exists der sit f(d) =0 A f(-M). HCM) -m x H M M fle) M fom ILM) for aro for a Lo In any case (azo or aco), foo has a real root :. The given statement is false. OR Every af f(x)=0 has no roots that is real, so it has all complex roots. But we know that complex roots occurs in pain. But & degree of f(x) is odd. So we get a contradiction. ... f(x)=0 has a real root. - The given statement is also.