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f ( x ) is a monic quadratic polynomial with positive real roots, and satisfies f ( 1 ) = 729 . What is the minimium value of f ( 0 ) ? Note: A monic polynomial has a leading coefficient of 1, i.e. f ( x ) = x n + … for some non-negative integer n .

f(x)is a monic quadratic polynomial with positive real roots, and satisfies f(1)=729. What is the minimium value of f(0)?


Note: A monic polynomial has a leading coefficient of 1, i.e. f(x)=xn+ for some non-negative integer n.

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

f(x) is a monic quadratic polynomial with positive real roots.
f(x) = x^{2} + bx + c
\Delta \geq 0
b^2 \geq 4c
f(1) = 1 + b + c
1 + b + c = 729
b + c = 728
b + \frac{b^2}{4} = 728
b^2 + 4b - 4(728) = 0
b^2 + 4b - 2912 = 0
b = -56, 52
c = \frac{b^2}{4}
c = \frac{52^2}{4}
c = 676
f(0)= c
f(0) = 676

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