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Marcosdiaz
@Marcosdiaz
April 2020
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Determine os pares ordenados de numeros reais (x,y) que representam as soluçoes do sistema
y+x=4
x²-xy-6=0
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DANILOCICILIOTTI
Verified answer
A) y= - x +4
quando y = 0 , x = 4 ( 4,0)
-> x = 4 ( colocando 0 em Y para obter X )
x²- xy - 6 = 0 quando y = 0 ( raiz6 , 0 )
x²- 0 - 6
x= +- raiz de 6
2 votes
Thanks 1
3478elc
y+x=4 ==> y = 4 - x
x²-xy-6=0
y1= 4 - 3 ==> y1 = 1
y2= 4 -(-1) ==> y2= 4+1 ==> y2=5
x^2 - (4-x)x - 6 = 0
x^2 +x^2 - 4x - 6 = 0
2x^2 - 4x - 6 =0 (:2)
x^2 - 2x - 3 = 0
delta= (-2)^2 - 4.1.(-3)==> 4 + 12 = 16
x=
2+/-V16
==>
2 +/- 4
2.1 2
x1=
2 + 4
==> x1 = 3
2
x2=
2 - 4
==> x2 = - 1
2
S = {(3, 1), (-1, 5)}
0 votes
Thanks 2
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Verified answer
A) y= - x +4quando y = 0 , x = 4 ( 4,0)
-> x = 4 ( colocando 0 em Y para obter X )
x²- xy - 6 = 0 quando y = 0 ( raiz6 , 0 )
x²- 0 - 6
x= +- raiz de 6
y+x=4 ==> y = 4 - x
x²-xy-6=0
y1= 4 - 3 ==> y1 = 1
y2= 4 -(-1) ==> y2= 4+1 ==> y2=5
x^2 - (4-x)x - 6 = 0
x^2 +x^2 - 4x - 6 = 0
2x^2 - 4x - 6 =0 (:2)
x^2 - 2x - 3 = 0
delta= (-2)^2 - 4.1.(-3)==> 4 + 12 = 16
x= 2+/-V16 ==> 2 +/- 4
2.1 2
x1= 2 + 4 ==> x1 = 3
2
x2= 2 - 4 ==> x2 = - 1
2
S = {(3, 1), (-1, 5)}