Question

[hoàng thiên]

Cho f(x) = x² + bx + c. Tìm b và c biết

a) f(1) = 2 ; f(–3) = 0

b) f(x) có nghiệm là 3 ; –6.

Answer 1

[Natsu Dragneel]

a) Nếu f(1) = 2, f(-3) = 0 ta có :
\(\left\{{}\begin{matrix}b+c+1=2\\9-3b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c=1\\3b-c=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4b=10\\b+c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\frac{5}{2}\\c=-\frac{3}{2}\end{matrix}\right.\)

b) Nếu f(x) có nghiệm là 3; -6 thì :

\(\left\{{}\begin{matrix}9+3b+c=0\\36-6b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3b+c=-9\\6b-c=36\end{matrix}\right.\)

\(\left\{{}\begin{matrix}9b=27\\6b-c=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\18-c=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\c=-18\end{matrix}\right.\)

Answer 2

[Buddy]

a,f(1)=2

⇔12+b.1+c=2(1)

f(−1)=0

⇔(−3)2−3b+c=0(2)

Từ: (1)(2) ta có hệ:

{b+c=1

−3b+c=−9

⇔{b=5\2c=−3\2

{b+c=1

−3b+c=−9

⇔{b=52

c=−32

f(x)=x2+52x−32

b,f(x)có n0là 3;−6

Ta có hệ pt: {3²+3b+c=0

(−6)²−6b+x=0

⇔{b=3

c=−18

ʃ x=x³+3x-18

⇒f(x)=x3+3x−18