find the valur of k for quadratic equation kx^2+(2k+4)x+9
Answers
Answered by
0
Answer:
4 and 1
Step-by-step explanation:
=4k^2+16+16k-k×9×4=0
=4k^2-20k+16=0
=k^2-5k+4=0
=k^2-4k-k+4=0
=K(k-4)-1(k-4)=0
=therefore K=1 and K=4
Answered by
0
Answer:
K=5
Step-by-step explanation:
P(x)=Kx^2+(2k+4)x+9=0
Kx^2+2kx+4x+9=0
Put x=-1
P(-1)=K(-1)^2+2k(-1)+4(-1)+9=0
K-2k-4+9=0
-k+5=0
-k=-5
K=5
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