Question

conta com baskara
c) x ao quadrado -6x+5=0
d) 5x ao quadrado -3 x=0
me ajudem pfv ​

Answer 1

Explicação passo-a-passo:

c)

x² -6x+5=0

a=1

b=-6

c=5

∆=b²-4.a.c

∆=(-6)²-4.(1).(5)

∆=36 -20

∆=16

x'=[ -( -6)+√16]/2.(1)

x'=[6 +4]/2

x'=[ 10]/2

x'=5

x"=[-(-6)-√16]/2.(1)

x"=[6 -4]/2

x"=2/2

x"= 1

S={( 1 ; 5)}

d)

5x² -3x=0

a=5

b=-3

c=0

∆=b²-4.a.c

∆=(-3)²-4.(5).(0)

∆=9

x'=[ -( -3)+√9]/2.(5)

x'= [3 +3]/10

x'= 6/10

x'=(6÷2)/(10÷2)

x'=3/5

x"=[-(-3)-√9]/2.(5)

x"=[ 3 -3]/10

x"= 0/10

x"= 0

S={( 0 ; 3/5)}

Answer 2

c)

$\Delta=b^2-4\cdot a\cdot c\\\Delta=(-6)^2-4\cdot 1\cdot 5\\\Delta=36-20\\\Delta=16\\ \\x=\frac{-b\pm\sqrt{\Delta}}{2\cdot a}\\ \\x=\frac{-(-6)\pm4}{2\cdot1}\\ \\x'=\frac{6+4}{2}=\frac{10}{2}=5\\ \\x''=\frac{6-4}{2}=\frac{2}{2}=1$

d)

$\Delta=b^2-4\cdot a\cdot c\\\Delta=(-3)^2-4\cdot5\cdot0\\\Delta=9\\ \\x=\frac{-b\pm\sqrt{\Delta} }{2\cdot a}\\ \\x=\frac{-(-3)\pm3}{2\cdot5}\\ \\x'=\frac{3+3}{5}=\frac{6}{10}\;(:2)=\frac{3}{5} \\ \\x''=\frac{3-3}{10}=\frac{0}{10}=0$

Bons estudos!