Question

Dla jakich wartości parametru m suma kwadratów pierwiastków równania jest równa 7 ?

Answer 1

x^2 - 2m x - 3 = 0

(x1)^2 + ( x2)^2 = ( x1 + x2)^2 - 2 x1*x2 = ( - b/a)^2 - 2*( c/a) = ( 2m)^2 - 2*( -3) =

= 4 m^2 + 6 = 7

4 m^2 = 1

m^2 = 1/4

m = - 1/2  lub  m = 1/2

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Answer 2

Warunki:

1) a ≥ 0

2) Δ ≥ 0

3) x1^2 + x2^2 = 7

 

1) a = 1 > 0 spełniony

2) Δ = (-2m)^2 - 4 * 1 * (-3) = 4m^2 + 12

    Δ ≥ 0 ⇔ 4m^2 + 12 ≥ 0

                 4m^2 ≥ - 12   /:4

                  m^2 ≥ -3 stąd m∈ R

 

3)x1^2 + x2^2 = x1^2 + 2x1x2 + x2 - 2x1x2 = (x1 + x2)^2 - 2x1x2 = (-b/a)^2 - 2*c/a =

(2m/1)^2 - 2*(-3)/1 = 4m^2 + 6

 

4m^2 + 6 = 7

4m^2 = 7 - 6

4m^2 = 1  /:4

m^2 = 1/4

m = 1/2 ∨ m = - 1/2