Question

The equation x^2-2kx + 1 = 0 has two distinct real roots. Find the set of all possible values of k.​

Answer 1

Answer: k < -1 or k > 1

Step-by-step explanation

The discriminant formula is

d = b^2 - 4ac

In this case we have

• a = 1
• b = -2k
• c = 1

If d > 0, then there are two distinct roots.

So,

d > 0

b^2 - 4ac > 0

(-2k)^2 - 4(1)(1) > 0

4k^2 - 4 > 0

4k^2 > 4

k^2 > 4/4

k^2 > 1

sqrt(k^2) > sqrt(1)

|k| > 1

k > 1 or k < -1

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Examples:

• When k = 5, the equation x^2-2kx+1 = 0 updates to x^2-10x+1 = 0. The two roots are x = 5 + 2*sqrt(6) and x = 5 - 2*sqrt(6)
• When k = 0, the equation x^2-2kx+1 = 0 updates to x^2+1 = 0 which doesn't have any real number roots. The two roots are complex numbers.
• When k = -3, the equation x^2-2kx+1 = 0 updates to x^2+6x+1 = 0. The two roots are x = -3 + 2*sqrt(2) and x = -3 - 2*sqrt(2)