X²+(k+1)2+k²=0 find value of k ( Quadratic Equation )
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Answered by
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Hello mate,
● Answer -
k = -1/2
● Explanation -
Comparing x²+ 2(k+1)x + k² = 0 with ax²+ bx + c = 0,
a = 1
b = 2(k+1)
c = k²
For real and equal roots,
b² - 4ac = 0
(2k+2)² = 4×1×k²
4k² + 8k + 4 = 4k²
8k + 4 = 0
k = -4/8
k = -1/2
Therefore, value of k must be -1/2 for given eqn to have real & equal roots.
Hope this helped...
● Answer -
k = -1/2
● Explanation -
Comparing x²+ 2(k+1)x + k² = 0 with ax²+ bx + c = 0,
a = 1
b = 2(k+1)
c = k²
For real and equal roots,
b² - 4ac = 0
(2k+2)² = 4×1×k²
4k² + 8k + 4 = 4k²
8k + 4 = 0
k = -4/8
k = -1/2
Therefore, value of k must be -1/2 for given eqn to have real & equal roots.
Hope this helped...
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