Question

Resolva, em IR, as equações do 2° Grau

A) 2x²-3x+1=0

B) -x²+2x+15=0

C) x²-5x+9=0

D) x²-3√3x+6=0

E) (3x-1)²+(x-2)²=25

F) 2.(x+3)²-5.(x+3)+2=0

G) x+1/x =3

Answer 1

Se o delta der negativo não a resultado real.

A) 2x^2 - 3x+1=0 a=2, b=-3, c=1

Delta:

${b}^{2} - 4ac \\ ( - 3) - 4.2.1 \\ 9 - 8 = 1 \\ \sqrt{1} = 1$

Baskara:

$\frac{ - b \frac{ + }{} \sqrt{delta} }{2a} \\ \frac{ - ( - 3) \frac{ + }{} 1}{2.2} = \frac{3 \frac{ + }{}1 }{4} \\ x1 = \frac{3 + 1}{4} = \frac{4}{4} = 1 \\ x2 = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}$

B) a= -1, b= 2, c= 15

Delta:

${b}^{2} - 4ac \\ {2}^{2} - 4.( - 1).15 \\ 4 + 60 = 64 \\ \sqrt{64} = 8$

Baskara:

$\frac{ - b \frac{ + }{} \sqrt{delta} }{2.a} \\ \frac{ - 2 \frac{ + }{}8 }{2.( - 1)} = \frac{ - 2 \frac{ + }{} 8}{ - 2} \\ x1 = \frac{ - 2 + 8}{ - 2} = \frac{6}{ - 2} = - 3 \\ x2 = \frac{ - 2 - 8}{ - 2} = \frac{ - 10}{ - 2} = 5$

C) a= 1, b= -3, c= 9

Delta:

${b}^{2} - 4ac \\ ( - 5) - 4.1.9 \\ 25 - 36 \\ - 11$

D) a= 1, b= -3 (raiz de 3), c= 6

Delta:

${b}^{2} - 4ac \\ ( - 3 \sqrt{3} ) - 4.1.6 \\ 9 \sqrt{9} - 24 \\ 9 + 3 - 24 \\ 12 - 24 \\ - 12$

E)

${(3x - 1)}^{2} + {(x - 2)}^{2} = 25 \\ (3x - 1).(3x - 1) + (x - 2).(x - 2) = 25 \\ 9 {x}^{2} - 3x - 3x + 1 + {x}^{2} - 2x - 2x + 4 - 25 \\ 9 {x}^{2} + {x}^{2} - 3x - 3x - 2x - 2x + 1 + 4 - 25 = 0 \\ 10 {x}^{2} - 10x - 20 = 0 \\ \frac{10 {x}^{2} - 10x - 20 }{10} \\ {x}^{2} - x - 2 = 0$

a= 1, b= -1, c= -2

Delta:

${b}^{2} - 4ac \\ {( - 1)}^{2} - 4.1.( - 2) \\ 1 + 8 = 9 \\ \sqrt{9} = 3$

Baskara:

$\frac{ - b \frac{ + }{} \sqrt{delta} }{2a} \\ \frac{ - ( - 1) \frac{ + }{} 3}{2.1} = \frac{1 \frac{ + }{} 3}{2} \\ x1 = \frac{1 + 3}{2} = \frac{4}{2} = 2 \\ x2 = \frac{1 - 3}{2} = \frac{ - 2}{2} = - 1$

F)

$2 {(x + 3)}^{2} - 5(x + 3) + 2 = 0 \\ {(2x + 6)}^{2} - 5x - 15 + 2 = 0 \\ (2x + 6).(2x + 6) - 5x - 13 = 0 \\ 4 {x}^{2} + 12x + 12x + 36 - 5x - 13 = 0 \\ 4 {x}^{2} + 12x + 12x - 5x + 36 - 13 = 0 \\ 4 {x}^{2} + 19x + 23 = 0$

a= 4, b= 19, c= 23

${b}^{2} - 4ac \\ {19}^{2} - 4.4.23 \\ 361 - 368 = - 7$

G)

$\frac{x + 1}{x} = \frac{3}{1} \\ 3x = x + 1 \\ 3x - x = 1 \\ 2x = 1 \\ x = \frac{1}{2}$