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If 2x square -(2+k)x +k=0 ,where k is a real no. ,find the roots of the equation.?
Most Upvoted Answer
If 2x square -(2+k)x +k=0 ,where k is a real no. ,find the roots of th...
By comparing the cofficient of given equation
2x²-(2+k)x+k=0
to standard quadratic equation
ax²+bx+c=0
a=2,b=-(2+k),c=k
Discriminant (D)=√b²-4ac
D=(-2-k)²-4×2×k
D=k²+4k+4
D=(k-2)²
by using quadratic formula
first Root=-b+√D/2a
first Root=k/2
second root=-b-√D/2a
=1
Community Answer
If 2x square -(2+k)x +k=0 ,where k is a real no. ,find the roots of th...
Solution:

Quadratic Equation:
The given quadratic equation is 2x^2 - (2+k)x + k = 0.

Roots of the Quadratic Equation:
To find the roots of the quadratic equation, we use the formula x = (-b ± √(b^2 - 4ac)) / 2a.

Identifying the Coefficients:
From the given equation, we have a = 2, b = -(2+k), and c = k.

Substitute the Coefficients:
Substitute the values of a, b, and c into the quadratic formula to get x = (-(2+k) ± √((2+k)^2 - 4*2*k)) / 2*2.

Simplify the Expression:
x = (-(2+k) ± √(4 + 4k + k^2 - 8k)) / 4.
x = (-(2+k) ± √(k^2 - 4k + 4)) / 4.

Calculate the Discriminant:
The discriminant, Δ = k^2 - 4k + 4.

Find the Roots:
1. If Δ > 0, then the equation has two distinct real roots.
2. If Δ = 0, then the equation has one real root.
3. If Δ < 0,="" then="" the="" equation="" has="" no="" real="" />

Conclusion:
By following the steps above and analyzing the discriminant, you can determine the nature of the roots of the given quadratic equation.
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If 2x square -(2+k)x +k=0 ,where k is a real no. ,find the roots of the equation.?
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