Question

find k,for which the quadratic equation (2k+1)x²-(7k+2)x+(7k-3)=0 has equal roots.Also find the roots.

Answer 1

answer is 4 or -4/3 . J equated the determinant to zero as its having equal roots.

Answer 2

a=(2k+1), b=-(7k+2), c=(7k-3)
d=b^2-4ac
=-(7k+2)^2-4(2k+1)(7k-3)
=49k^2+4+28k-4(14k^2+7k-6k-3)
=49k^2+4+28k-56k^2-28k+24k+12
=-7k^2+24k+16
=0 (Because the roots are real and equal)
-7k^2+24k+16=0
7k^2-24k-16=0
7k^2-28k+4k-16=0
7k(k-4)+4(k-4)=0
(7k+4)(k-4)=0
k=-4/7 or 4
roots:
-b/2a=(7k+2)/2(2k+1)
=30/18