★ 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
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
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