Question

xsquare-2(k+1)x+ksquare =0 has equal root​

Answer 1

$x {}^{2} - 2(k + 1)x + k {}^{2} = 0$

For equal root D = 0

$b {}^{2} - 4ac = 0$

$(2k + 2) {}^{2} - 4(1)k {}^{2} = 0$

$4k {}^{2} + 8k + 4 - 4k {}^{2} = 0$

$8k = - 4$

$k = - 4 \div 8 =$

K = - 1/2

Thanks

Answer 2

Solution '

Compare given Quadratic

equation x²-2(k+1)x+k²=0

by ax²+bx+c= 0, we get

a = 1 , b = -2(k+1), c = k²

Discreminant (D)= 0

[ Given equal roots ]

=> b² - 4ac = 0

=> [-2(k+1)]² -4×1×k² = 0

=> [2(k+1)]²-(2k)² = 0

=>[2(k+1)+2k][2(k+1)-2k]=0

2(k+1)+2k=0or2(k+1)-2k=0

=>2k+2+2k=0or 2k+2-2k=0

=> 4k+2 = 0

=> 4k = -2

=> k = -2/4

=> k = -1/2

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