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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)=xn+… for some non-negative integer n.

makarand gomashe , 12 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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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