Question

EQUAÇÃO DE 2° GRAU:

1)
$x {}^{2} - x - 20 = 0$
2)
$x {}^{2} - 7x + 12 = 0$
3)
$3y {}^{2} + 2y - 1 = 0$
4)
$x {}^{2} + 6x + 9 = 0$
5)
$9x {}^{2} - 6x + 9 = 0$
6)
$- 3t {}^{2} + 1 + 4 = 0$
7)
${x}^{2} - 2x - 1 = 0$
8)
$6y {}^{2} + y - 1 = 0$
9)
$u {}^{2} + 4u - 5 = 0$

10)
$- 16x {}^{2} + 8x - 1 = 0$

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Answer 1

Explicação passo-a-passo:

1) x² - x -20=0

∆= (-1)² -4.1.(-20)

∆= 1 + 80

∆= 81

x = - (-1) ±√81/2.1

x= 1 ±9/2

x1= 1+9/2= 10/2= 5

x2= 1-9/2= -8/2= -4

S = { -4, 5}

2) x² -7x +12=0

∆= (-7)² -4.1.12

∆= 49 - 48

∆= 1

x= - (-7) ±√1/2.1

x= 7 ±1/2

x1= 7+1/2= 8/2=4

x2= 7-1/2= 6/2=3

S= { 3, 4}

3) 3y² +2y - 1=0

∆= 2² - 4.3.(-1)

∆= 4 + 12

∆= 16

y= - 2 ±√16/2.3

y= - 2 ±4/6

y1= -2+4/6= 2/6= 1/3

y2= -2 -4/6= -6/6= -1

S= { -1, 1/3}

4) x² +6x +9=0

∆= 6² - 4.1.9

∆= 36 - 36

∆= 0

x= - 6±√0/2.1

x= -6/2

x= -3

S = { - 3}

5) 9x² -6x +9=0

∆= (-6)² -4.9.9

∆= 36 - 324

∆= - 288

Não existe raiz quadrada de número negativo

S = { } vazio

6) se for:

- 3t²+1t +4=0

3t² - t - 4=0

∆= (-1)² - 4.3.(-4)

∆= 1 + 48

∆= 49

t = - (-1) ±√49/2.3

t= 1 ±7/6

t1= 1+7/6= 8/6= 4/3

t2= 1-7/6= -6/6= -1

S = { - 1, 4/3}

7) x² - 2x -1=0

∆= (-2)² -4.1.(-1)

∆= 4 + 4

∆= 8

x = - (-2) ±√8/2.1

x= 2 ± √8/2

*8= 2.2.2= 2².2

√8= √2².2= 2√2

x= 2 ±2√2/2

x1= 1 + √2

x2= 1 - √2

S = { 1+√2 , 1-√2}

8) 6y² +y -1=0

∆= 1² - 4.6 .(-1)

∆= 1 + 24

∆= 25

y= - 1 ±√25/2.6

y= - 1 ±5/12

y1= -1+5/12= 4/12= 1/3

y2= -1-5/12= -6/12= - 1/6

S = { -1/6, 1/3}

9) u² +4u -5=0

∆= 4² -4.1.(-5)

∆= 16 + 20

∆= 36

u= - 4 ±√36/2.1

u= -4 ±6/2

u1= -4 +6/2= 2/2=2

u2= -4-6/2= -10/2= -5

S = { -5, 2}

10)-16x² +8x-1=0

16x² -8x +1= 0

∆= (-8)² - 4.16.1

∆= 64 - 64

∆= 0

x = -(-8)/2.16

x= +8/32

x= 1/4

S = { 1/4}