Alan

General quadratic polynomial:  P(x) = ax^2 + bx + c

P(0) = 1   ⇒   1 =  c

P(1) = 5   ⇒   5 = a + b + c               so 5 = a + b + 1            or   4 = a + b   (1)

P(2) = 11  ⇒  11 = 4a + 2b + c          so 11 = 4a + 2b + 1      or  10 = 4a + 2b  (2)

Multiply (1) by 2 and subtract from from (2) to get:    2 = 2a  or a = 1

Put this back into (1) to get:  4 = 1 + b  or b = 3

So:  P(x) = x^2 + 3x + 1

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What is your question?

Rearrange the equation to look like:

x^2 -2(2k+1)x + k+2 = 0

This has equal roots when the discriminant is zero.

(In general, when a*x^2 + b*x + c = 0, the discriminant is b^2 - 4*a*c): 

(-2(2k + 1))^2 - 4*1*(k+2) = 0

4*(2k+1)^2 - 4*(k+2) = 0

(2k+1)^2 - (k+2) = 0

4k^2 + 4k + 1 - k - 2 = 0

4k^2 + 3k - 1 = 0

(4k - 1)(k + 1) = 0

4k = 1 or k = -1 

k = 1/4 or k = -1

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What do you want to do with this?  Plot it?

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It's the equation of a straight line whose slope is -2/3 and whose y-intercept is -40/3

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As follows:

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With the factorial sign, 3! means 3*2*1 or just 3*2, by definition.

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18C3= 18!/(15!*3!) = (18*17*16)/(3*2)   Note that the denominator has 3! not just 3  (3! = 3*2)

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This is a well known problem called the "Monty Hall" problem.  Have a look at https://en.m.wikipedia.org/wiki/Monty_Hall_problem for example.

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