Find the range of values of k for which the equation x^2+5kx+16=0
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Answered by
0
f (x)=x^2+5kx+16
f (1)=1^2+5×1×k+16
=17+5k=0
=k=-17/5.
f (1)=1^2+5×1×k+16
=17+5k=0
=k=-17/5.
Answered by
1
b^2-4ac<0
25k^2-64<0
k^2<64/25
-8/5<k<8/5
25k^2-64<0
k^2<64/25
-8/5<k<8/5
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