roots of xsquare +2x+k=0 are complex then a possible value of k is
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Answered by
1
If x²+2x+k=0 has complex roots, then its determinant (D) is less than 0.
D=b²-4ac
b²-4ac<0
2²-4×1×k<0
4-4k<0
4k>4
k>1
k={2,3,4,5,6,....}
D=b²-4ac
b²-4ac<0
2²-4×1×k<0
4-4k<0
4k>4
k>1
k={2,3,4,5,6,....}
Answered by
3
Hi ,
Compare x² + 2x + k = 0 with
ax² + bx + c = 0, we get
a = 1 , b = 2 , c = k ,
Discreaminant ( D ) < 0
[Given Roots are complex ]
D = b² - 4ac < 0
2² - 4 × 1 × k < 0
4 - 4k < 0
- 4k < - 4
k > ( - 4 ) /( - 4 )
k > 1
I hope this helps you.
: )
Compare x² + 2x + k = 0 with
ax² + bx + c = 0, we get
a = 1 , b = 2 , c = k ,
Discreaminant ( D ) < 0
[Given Roots are complex ]
D = b² - 4ac < 0
2² - 4 × 1 × k < 0
4 - 4k < 0
- 4k < - 4
k > ( - 4 ) /( - 4 )
k > 1
I hope this helps you.
: )
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