a) Thay \(k=0\) vào phương trình ta có:
\(4x^2-25+0^2+4.0.x=0\)
\(⇔4x^2-25=0\)\(⇔(2x)^2-5^2=0\)
\(⇔(2x-5)(2x+5)=0\)
\(⇔\left[\begin{array}{} 2x-5=0\\ 2x+5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{5}{2}\\ x=-\frac{5}{2} \end{array}\right.\)
b) Thay \(k=-3\) vào phương trình ta có:
\(4x^2-25+(-3)^2+4.(-3).x=0\)
\(⇔4x^2-12x-16=0\)
\(⇔4x^2-16x+4x-16=0\)
\(⇔4x(x-4)+4(x-4)=0\)
\(⇔(x-4)(4x+4)=0\)
\(⇔\left[\begin{array}{} x-4=0\\ 4x+4=0 \end{array}\right.⇔\left[\begin{array}{} x=4\\ x=-1 \end{array}\right.\)
c) Thay \(x=-2\) vào pt ta có:
\(4.(-2)^2-25+k^2+4k.(-2)=0\)
\(⇔k^2-8k-9=0\)
\(⇔k^2-9k+k-9=0\)
\(⇔k(k-9)+(k-9)=0\)
\(⇔(k-9)(k+1)=0\)
\(⇔\left[\begin{array}{} k-9=0\\ k+1=0 \end{array}\right.⇔\left[\begin{array}{} k=9\\ k=-1 \end{array}\right.\)
