Question

Dla jakiego "m" równanie ma dwa pierwiastki, których suma kwadratów wynosi 400

Answer 1

a = 1    b = 5m    c = 20m - 8
1. Δ ≥ 0
2. x1^2 + x2^2 = 400

ad. 1
Δ =(5m)^2 - 4 * 1 * (20m - 8) = 25m^2 - 80m + 32
25m^2 - 80m + 32 ≥ 0
 Δ (m) = 6 400 - 4 * 25 * 32 = 6 400 - 3 200 = 3 200
√Δ = 40√2
m1 = (80 - 40√2) / 50 = (8 - 4√2)/5
m2 = (80 + 40√2) / 50 = (8 + 4√2)/5
m należy (-oo, m1>  u <m2 ,+oo)

ad. 2
x1^2 + x2^2 =4 00
(x1 + x2)^2 - 2x1x2 = 400
(-b/a)^2  - 2*c/a = 400
(-5m /1)^2 - 2*(-8/1 )= 400
25m^2 + 16 = 400
25m^2 = 384           /:25
m^2 = 384/25
m = √384/25   lub m  = -√384/25
m = 8√6 / 5            m = -8√6 / 5

odp. m = 8√6 / 5            m = -8√6 / 5

Answer 2

D=>0
(x1)^2+(x2)^2=400
delta=25m^2-4*1*(20m-8)=25m^2-80m+32
25m^2-80m+32=>
mała delta= 6400-4*25*32=6400-3200=3200
pierw z małej delty=40 pierw z 2
m1=(80-40pierw z 2)/2*25=(80-40 pierw z 2)/50= (8-4 pierw z 2)/5
m2=(8+4pierw z 2)/5
D:me(-nieskoń, (8-4 pierw z 2)/5> U<(8+4 pierw z 2)/5 ,nieskoń)

(x1)^2+(x2)^2=400
(x1+x2)^2-2x1x2=400
(-b/a)^2 -2*c/a=400
(-5m /1)^2 -2*(-8/1)=400
25m^2+16=400
25m^2=384 /:25
m^2=384/25
m= -pierw z 384/25=( -8pierw z 6)/5 lub m= pierw z 384/25= (8 pierw z 6)/5
(8 pierw z 6)/5 eD
(- 8pierw z 6)/5 eD