Question

Cho đa thức P(x)=\(^{x^2}\)+ax+b
P(1)=3;P(-2)=4 tìm a,b

Answer 1

Ta có: \(P\left(1\right)=1^2+a\cdot1+b=3=a+b+1=3\Leftrightarrow a+b=2\)(1)

\(P\left(-2\right)=\left(-2\right)^2+\left(-2\right)a+b=4-2a+b=4\Rightarrow b-2a=0\)

\(\Rightarrow\left(a+b\right)-\left(b-2a\right)=2-0=2\)

\(\Rightarrow a+b-b+2a=2\)

\(\Rightarrow3a=2\Rightarrow a=\dfrac{2}{3}\)

Thay \(a=\dfrac{2}{3}\) vào (1) ta được:

\(\dfrac{2}{3}+b=2\Rightarrow b=2-\dfrac{2}{3}=\dfrac{6}{3}-\dfrac{2}{3}=\dfrac{4}{3}\)

Vậy \(a=\dfrac{2}{3};b=\dfrac{4}{3}\)

Answer 2

p(1) = 1 => 1+ a + b = 1 => a + b = 0

p(-1) = 3 => 1 - a + b = 3 => b - a = 4

=> a = (0 - 4 ) : 2 = -2

b = (0 + 4) : 2 = 2