Question

[Trần Ngọc Thảo]

Cho f(x) = x² + bx + c

a) f(1) = 2 ; f(-3) = 0

b) f(x) có nghiệm là 3 ; -6

Answer 1

[Diệu Huyền]

\(a,f\left(1\right)=2\)

\(\Leftrightarrow1^2+b.1+c=2\left(1\right)\)

\(f\left(-1\right)=0\)

\(\Leftrightarrow\left(-3\right)^2-3b+c=0\left(2\right)\)

Từ: \(\left(1\right)\left(2\right)\) ta có hệ:

\(\left\{{}\begin{matrix}b+c=1\\-3b+c=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\frac{5}{2}\\c=-\frac{3}{2}\end{matrix}\right.\)

\(f\left(x\right)=x^2+\frac{5}{2}x-\frac{3}{2}\)

\(b,f\left(x\right)\) có \(n_0\) là \(3;-6\)

Ta có hệ pt: \(\left\{{}\begin{matrix}3^2+3b+c=0\\\left(-6\right)^2-6b+x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\c=-18\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)=x^3+3x-18\)