Question

(x)2-2kx+7k-12=0the value of k such that the equation has real and equal roots k
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Answer 1

Answer:

Comparying x^2-2kx+7k-12 with ax^2+bx+c,

a=1; b= -2k; c=7k-12

As, the equation has real and equal roots,

b^2-4ac=0

=>(-2k)^2-4×1(7k-12)=0

=>4k^2-28k+48=0

=>4(k^2-7k+12)=0

=>k^2-(3k+4k)+12=0

=>k(k-3)-4(k-3)=0

=>(k-3)(k-4)=0

Either,

k-3=0 or k-4=0

=>k=3 =>k=4