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What is the value of n in the equation of 1/n=x^2-x+1

if the roots are unequal and real
n>0

User Gct
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Answer:

Hope this helps and have a nice day

Explanation:

To find the value of n in the equation 1/n = x^2 - x + 1, given that the roots are unequal and real, and n > 0, we can analyze the properties of the equation.

The equation 1/n = x^2 - x + 1 can be rearranged to the quadratic form:

x^2 - x + (1 - 1/n) = 0

Comparing this equation to the standard quadratic equation form, ax^2 + bx + c = 0, we have:

a = 1, b = -1, and c = (1 - 1/n).

For the roots of a quadratic equation to be real and unequal, the discriminant (b^2 - 4ac) must be positive.

The discriminant is given by:

D = (-1)^2 - 4(1)(1 - 1/n)

= 1 - 4 + 4/n

= 4/n - 3

For the roots to be real and unequal, D > 0. Substituting the value of D, we have:

4/n - 3 > 0

Adding 3 to both sides:

4/n > 3

Multiplying both sides by n (since n > 0):

4 > 3n

Dividing both sides by 3:

4/3 > n

Therefore, for the roots of the equation to be unequal and real, and n > 0, we must have n < 4/3.

User Dave Lillethun
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