Question

4x2-25+k2+4kx=0 a)gpt với k=0 b) tìm k để pt nhận x= -2 là nghiệm

Answer 1

a,Thay k=0 vào pt ta có:
\(4x^2-25+0^2+4.0.x=0\\ \Leftrightarrow4x^2-25=0\\ \Leftrightarrow x^2=\dfrac{25}{4}\\ \Leftrightarrow x=\pm\dfrac{5}{2}\)

b, Thay x=-2 vào pt ta có:
\(4.\left(-2\right)^2-25+k^2+4k.\left(-2\right)=0\\ \Leftrightarrow4.4-25+k^2-8k=0\\ \Leftrightarrow k^2-8k+16-25=0\\ \Leftrightarrow k^2-8k-9=0\\ \Leftrightarrow\left(k^2+k\right)-\left(9k+9\right)=0\\ \Leftrightarrow k\left(k+1\right)-9\left(k+1\right)=0\\ \Leftrightarrow\left(k+1\right)\left(k-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}k=-1\\k=9\end{matrix}\right.\)