Math, asked by LaRouge, 1 month ago

★ Find the value of k. So that, the Quadratic equation have real roots => y² + k² = 2(k+1)y ㅤ
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Answers

Answered by nirmaladevirnc1985
2

Step-by-step explanation:

Nature of the roots of a quadratic equation is determined by its discriminant D=b

2

−4ac

Comparing x

2

−2(k+1)x+k

2

=0 with ax

2

+bx+c=0

we get a=1,b=−2(k+1),c=k

2

Therefore for equal roots,

D=b

2

−4ac=0

=>[−2(k+1)]

2

−4×1×k

2

=0

=>4(k

2

+2k+1)−4k

2

=0

=>4k

2

+8k+4−4k

2

=0

=>8k+4=0

=>k=−

8

4

=>k=−

2

1

Answered by shinau993
0

Correct option is C)

Nature of the roots of a quadratic equation is determined by its discriminant D=b

2

−4ac

Comparing x

2

−2(k+1)x+k

2

=0 with ax

2

+bx+c=0

we get a=1,b=−2(k+1),c=k

2

Therefore for equal roots,

D=b

2

−4ac=0

=>[−2(k+1)]

2

−4×1×k

2

=0

=>4(k

2

+2k+1)−4k

2

=0

=>4k

2

+8k+4−4k

2

=0

=>8k+4=0

=>k=−

8

4

=>k=−

2

1

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