What is the value of c? That is the question.

Algebra Level 3

Knowing f ( x ) = x 2 + a x + b f(x) = x^2 + ax + b for real a a and b b has a range of [ 0 , ∞ ) [0, \infty ) , and f ( x ) < c f(x) < c has a solution set ( m , m + 6 ) (m, m+6) , what is the value of c c ?


The answer is 9.

Kevin Xu by Kevin Xu , 19, Canada

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1 solution

Kevin Xu
Aug 27, 2019

∵ f ( x ) \because f(x) has range [ 0 , ∞ ) [0, \infty) , thus discriminant Δ = 0 − − > a 2 = 4 b \Delta = 0 --> a^2 = 4b \\

∴ x 2 + a x + a 2 4 − c < 0 \therefore x^2 +ax + \frac {a^2}{4}-c<0 has solution set ( m , m + 4 ) (m, m+4) \\ \\ Using Vieta's Formula we get: \\ { 2 m + 6 = − 1 m ( m + 6 ) = a 2 4 − 4 \begin{cases} 2m + 6 = -1 \\m(m+6)=\frac {a^2}{4}-4 \end{cases} \\

Solve and get c = 9 c = 9

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