# Given M=minim of sum of (x_k)^2 where x_k are roots of equation x^2+(m-2)x-(m+3)=0, then: a) k=1; b) k=2; c) k=9; d)k=6 ; e) k=1/2; f) k=8

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### 1 Answer

The answer provided by the problem needs to have a connection with the minimum value required, hence, you should replace M for k. This assumption is made based on the logical reasoning regarding the request of the problem and the information provided by the problem. The value of k, cannot be larger than 2, since the quadratic equation cannot have more than 2 roots. You need to evaluate the summation of the squares of roots of the given quadratic equation, such that:

`x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1*x_2`

Using Vieta's equations, yields:

`x_1 + x_2 = -b/a`

`x_1*x_2 = c/a`

Identifying the coefficients of quadratic equation, yields:

`a = 1, b = m - 2, c = -(m + 3)`

`x_1 + x_2 = -b/a => x_1 + x_2 = 2 - m`

`x_1*x_2 = c/a => x_1*x_2 = -(m + 3)`

Replacing `2 - m` for `x_1 + x_2 ` and `-(m + 3)` for `x_1*x_2` yields:

`x_1^2 + x_2^2 = (2 - m)^2 - 2(-(m + 3))`

`x_1^2 + x_2^2 = 4 - 4m + m^2 + 2m + 6`

`x_1^2 + x_2^2 = m^2 - 2m + 10`

You need to evaluate the minimum of summation of squares of the roots, hence, you need to differentiate the quadratic `f(m) = m^2 - 2m + 10` with respect to m, such that:

`f'(m) = 2m - 2`

You need to solve for m the equation `f'(m) = 0` , to find the minimum, such that:

`2m - 2 = 0 => 2m = 2 => m = 1`

**Hence, evaluating the minimum of summation of squares of the roots yields `M = 1` , thus, you need to select the answer **`a) M = 1.`