Find the value of k for which the equation (k-5)x
^2+2(k-5)x+2=0 has real
and equal roots.
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chinmayeepriya:
tysm^-^
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we have the equation have real and equal roots so b2-4ac=0 now
2(k-5)whole square-4.2(k-5)=0 ,4ksquare-40k+100-8k+40=0. again 16ksquare-48k+140=0 also we can write as ksquare-12k+35=0now ksquare -(7+5)k+35=0,soksquare-7k-5k+35=0. therefore k=7ork=5
2(k-5)whole square-4.2(k-5)=0 ,4ksquare-40k+100-8k+40=0. again 16ksquare-48k+140=0 also we can write as ksquare-12k+35=0now ksquare -(7+5)k+35=0,soksquare-7k-5k+35=0. therefore k=7ork=5
:) tysm
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