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2014-08-15T20:50:07-03:00
9x - x² = -22
-x²+9x+22=0
a= -1
b= 9
c= 22

Δ=b²-4.a.c
Δ=9²-4.-1.22
Δ=81+88
Δ=169

x= -b+-√Δ

x' = -9+13/2.-1  = 4/-2 = -2

x'' = -9-13/2.-1   = -22/-2 = 11

S= ( -2,11 )


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  • Usuário do Brainly
2014-08-15T21:53:39-03:00
Seja j o número procurado.

Assim, 9j-j^2=-22.

Deste modo, j^2-9j-22=0.

Temos \Delta=(-9)^2-4\cdot1\cdot(-22)=169.

Logo, j=\dfrac{9\pm\sqrt{169}}{2}=\dfrac{9\pm13}{2}.

Portanto, j'=\dfrac{9+13}{2}=11 e j''=\dfrac{9-13}{2}=-2.