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Question

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802

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Given x_1,x_2 are the real roots of x^2-2mx+(m^2+2m+3)=0, please determine the minimum value of (x_1)^2+(x_2)^2.

yasbib555 Oct 5, 2018

edited by yasbib555 Dec 9, 2018

CPhill · Answer 1 · 2018-10-05T11:14:49

Let

x1 = [( 2m + √ [ 4m^2 - 4(1)(m^2 + 2m + 3 ) ] ) / 2 ]

x2 = [( 2m - √ [ 4m^2 - 4(1)(m^2 + 2m + 3 ) ] ) / 2 ]

(x1)^2 + (x2)^2 =

2 * [4m^2 + 4m^2 - 4m^2 - 8m - 12 ] / 4 =

[4m^2 - 8m - 12 ] / 2 =

2m^2 - 4m - 6

The m value that minimizes this is 4 / (2 * 2) = 4 / 4 = 1

So...the minimum value of (x1)^2 + (x2)^2 = 2(1)^2 - 4(1) - 6 = -8