if the sum of the roots pf the quadratic equation 2x2+(2k-1)x - (k-4)=0
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Compare given Quadratic equation
3x²+(2k+1)x-(k+5) = 0 with
ax²+bx+c=0, we get
a = 3 , b = (2k+1), c = -(k+5)
i ) Sum of the roots = -b/a
= -(2k+1)/3 ---(1)
ii) product of the roots = c/a
= -(k+5)/3 ----(2)
According to the problem given,
-(2k+1)/3 = -(k+5)/3
=> 2k+1 = k+5
=> 2k-k = 5-1
=> k = 4
Therefore,
k = 4.
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