Question

me ajudem a resolver essa primera questao pf ​

Answer 1

Resposta:

Explicação passo a passo:

$a)$

$3x^{2} -5x+1=0$

$a=3\\b=-5\\c=1$

Δ = $b^{2} -4.a.c$

Δ = $(-5)^{2} -4.(3).(1)$

Δ = $25-12$

Δ = $13$

x = ( - b ± √Δ ) / 2a

x = ( 5 ± √13 ) / 6

x' = ( 5 + √13 ) / 6

x'' = ( 5 - √13 ) / 6

S = { ( 5 + √13 ) / 6  ;  ( 5 - √13 ) / 6 }

$b)$

$-x^{2} +x=5$

$-x^{2} +x-5=0 .(-1)$

$x^{2} -x+5=0$

$a=1\\b=-1\\c=5$

Δ = $b^{2} -4.a.c$

Δ = $(-1)^{2} -4.(1).(5)$

Δ = $1-20$

Δ = $-19$

Como não existe raiz quadrada de n° negativo, essa equação

não tem solução.

S = ∅

$c)$

$x^{2} -4=0$

$x^{2} =4$

$x=$ ± $\sqrt{4}$

$x=$ ± $2$

x' =  $2$

x'' = $-2$

S = { - 2 ; 2 }

$d)$

$-3x^{2} -2x+5=0.(-1)$

$3x^{2} +2x-5=0$

$a=3\\b=2\\c=-5$

Δ = $b^{2} -4.a.c$

Δ = $2^{2} -4.3.(-5)$

Δ = $4+60$

Δ = $64$

x = ( - b ± √Δ ) / 2a

x = ( - 2 ± √64 ) / 6

x = ( - 2 ± 8 ) / 6

x' = ( - 2 + 8 ) / 6

x' = 6 / 6

x' = 1

x'' = ( - 2 - 8 ) / 6

x'' = ( - 10 ) / 6

x'' = - 5 / 3

S = { - 5 / 3  ;  1 }

$e)$

$4x^{2} +x-5=0$

$a=4\\b=1\\c=-5$

Δ = $b^{2} -4.a.c$

Δ = $1^{2} -4.4.(-5)$

Δ = $1+80$

Δ = $81$

x = ( - b ± √Δ ) / 2a

x = ( - 1 ± √81 ) / 8

x = ( - 1 ± 9 ) / 8

x' = ( - 1 + 9 ) / 8

x' = 8 / 8

x' = 1

x'' = ( - 1 - 9 ) / 8

x'' = ( - 10 ) / 8

x'' = - 5 / 4

S = { - 5 / 4  ;  1 }

$f)$

$-x^{2} +2x-1=0.(-1)$

$x^{2} -2x+1=0$

$a=1\\b=-2\\c=1$

Δ = $b^{2} -4.a.c$

Δ = $(-2)^{2} -4.1.1$

Δ = $4-4$

Δ = $0$

x'' = x'' = - b / 2a

x' = x'' = 2 / 2

x' = x'' = 1

S = { 1 }

$g)$

$-x^{2} +6x-9=0.(-1)$

$x^{2} -6x+9=0$

$a=1\\b=-6\\c=9$

Δ = $b^{2} -4.a.c$

Δ = $(-6)^{2} -4.1.9$

Δ = $36-36$

Δ = $0$

x'' = x'' = - b / 2a

x' = x'' = 6 / 2

x' = x'' = 3

S = { 3 }

Ufa!!!