Question

for what value of k are the roots of the quadratic equation y^2+k^2=2(k+1)t equal plz ans fastly

Answer 1

Here is solution for your question bro...
Y2 + K2 - 2 ( K - 1 ) Y = 0

Y2 - 2( K-1 ) Y + K2 = 0

A = 1, B = -2( K - 1) , C = K2

GIVEN ROOTS ARE EQUAL, THEN B2 - 4AC = 0

{ - 2 ( K - 1 ) } SQUARE - 4 * 1 * K2 = 0

4 ( K2 - 2K + 1) - 4K2 = 0

4 { K2 - 2K + 1 - K2} = 0

4 { 1 - 2K } = 0

I - 2K = 0/4 = 0

1 = 2K

1/2 = K
K=1/2

Answer 2

Y2 + K2 - 2 ( K - 1 ) Y = 0

Y2 - 2( K-1 ) Y + K2 = 0

A = 1, B = -2( K - 1) , C = K2

GIVEN ROOTS ARE EQUAL, THEN B2 - 4AC = 0

{ - 2 ( K - 1 ) } SQUARE - 4 * 1 * K2 = 0

4 ( K2 - 2K + 1) - 4K2 = 0

4 { K2 - 2K + 1 - K2} = 0

4 { 1 - 2K } = 0

I - 2K = 0/4 = 0

1 = 2K

1/2 = K

K=1/2