\(Ta\) \(có\)
\(P_{\left(1\right)}\)\(=\) \(8\) \(\Rightarrow\) \(a^2+2.b.1-3\)\(=\) \(8\)
\(\Rightarrow\) \(a^2+2b=11\) \(_{\left(1\right)}\)
\(P_{\left(-2\right)}\) \(=\) \(5\) \(\Rightarrow\) \(a^2+2.b.\left(-2\right)-3\) \(=\) \(5\)
\(\Rightarrow\) \(a^2-4b=8\) \(_{\left(2\right)}\)
\(Từ\) \(_{\left(1\right)}\)\(,\)\(_{\left(2\right)}\) \(trừ\) \(hai\) \(vế\) \(cho\) \(nhau\) \(ta\) \(được\)
\(\left(a^2+2b\right)-\left(a^2-4b\right)=11-8\)
\(\Rightarrow\)\(a^2+2b-a^2+4b=3\)
\(\Rightarrow\)\(6b=3\)
\(\Rightarrow\) \(b=\dfrac{1}{2}\)
\(Thay\) \(b=\dfrac{1}{2}\) \(vào\) \(_{\left(1\right)}\)\(ta\) \(có\)
\(a^2+2b=11\)
\(\Rightarrow\)\(a^2+2.\dfrac{1}{2}=11\)
\(\Rightarrow\) \(a^2+1=11\)
\(\Rightarrow\) \(a^2=10\)
\(\Rightarrow\) \(a=\sqrt{10}\)
\(Vậy\) \(a=\sqrt{10}\) \(và\) \(b=\dfrac{1}{2}\)
P(x)=a2+2bx-3
P(1)=a2+2b.1-3
=a2+2b-3
mà P(1)=8
=> a2+2b-3=8
a2+2b=11
P(-2)=a2+2b(-2)-3
= a2-4b-3
mà P(-2)=5
=>a2-4b-3=5
a2-4b=8
=>(a2-4b)-(a2+2b)=8-5
a2-4b-a2-2b=3
-6b=3
b=-1/2
=>a2-2=8
a2=10
=>a=\(\sqrt{10}\)
