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The quartic x^4 - 8x^3 + a x^2 - b x + 16 = 0 has four positive real…
Question
The quartic x^4 - 8x^3 + a x^2 - b x + 16 = 0 has four positive real roots (counted with multiplicity), where a and b are real constants. Find a + b.
✓ Verified answer: 56checked by our engine — not a guess
Step-by-step solution
Let the four positive roots be r1,r2,r3,r4. By Vieta, their sum is 8 and their product is 16. By AM-GM,
(r1+r2+r3+r4)/4 ≥ (r1 r2 r3 r4)^{1/4}, i.e. 8/4 = 2 ≥ 16^{1/4} = 2.
Equality in AM-GM forces all roots equal, so r1=r2=r3=r4 = 2 and the polynomial is (x-2)^4.
Expanding (x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16, we read off a = 24 and b = 32 (since -b = -32). Therefore a + b = 56.
Final answer56
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