Question

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1) Considere que X e R e resolva as seguintes equações:

a) x²-14x+45=0

b)x²-11x+30=2

c) x²-6x+5=-3

d) x²+12x+36=16

e) x²=x+6.

f) x²-5x+6=-x²+x+42

g) (5-x)²=(2x-1)²

Answer 1

A
${x}^{2} - 14x + 45 \\ delta = {14}^{2} - 4 \times 1 \times 45 \\ delta = 196 - 180 \\ delta = 16 \\ x1 = \frac{14 + 4}{2} \: x1 = 9 \\ x2 = \frac{14 - 4}{2} \: x2 = 5$
B
${x}^{2} - 11x + 30 = 2 \\ {x}^{2} - 11x + 28 = 0 \\ \\ delta = {11}^{2} - 4 \times 1 \times 28 \\ delta = 121 - 112 \\ delta = 9 \\ \\ x1 = \frac{11 + 3}{2} = 7 \\ x2 = \frac{11 - 3}{2} = 4$

C
${x}^{2} - 6x + 5 = - 3 \\ {x}^{2} - 6x + 8 = 0 \\ \\ delta = {6}^{2} - 4 \times 1 \times 8 \\ delta = 36 - 32 \\ delta = 4 \\ \\ x1 = \frac{6 + 2}{2} = 4 \\ \\ x2 = \frac{6 - 2}{2} = 2$
D
${x}^{2} + 12x + 36 = 16 \\ {x }^{2} + 12 + 10 \\ \\ delta = {12}^{2} - 4 \times 1 \times 10 \\ delta = 144 - 40 \\ delta = 104(delta \: inesistente)$
E
${x}^{2} = x + 6 \\ {x}^{2} - x - 6 = 0 \\ delta = - {1}^{2} - 4 \times 1 \times - 6 \\ delta = 1 + 24 \\ delta = 25 \\ \\ x1 = \frac{1 + 5}{2} = 3 \\ \\ x2 = \frac{1 - 5}{2} = - 2$
F
${x}^{2} - 5x + 6 = {x}^{2} + x + 42 \\ 2 {x}^{2} - 6x - 36 = 0 \\ \\ delta = {6}^{2} - 4 \times 2 \times - 36 \\ delta = 36 + 288 \\ delta = 324 \\ \\ x1 = \frac{6 + 18}{4} = 6 \\ \\ x2 = \frac{6 - 18}{4} = - 3$
G
${(5 - x)}^{2} = {(2x - 1)}^{2} \\ (5 - x) \times (5 - x) = (2x - 1) \times(2x - 1) \\ 25 - 10x \times {x}^{2} = {4x}^{2} - 4x + 1 \\ { - 3x}^{2} + 6x - 24 \\ delta = {6}^{2} - 4 \times 3 \times - 24 \\ delta = 36 + 288 \\ delta = 324 \\ \\ x1 = \frac{6 + 18}{6} = 4 \\ \\ x2 = \frac{6 - 18}{6} = - 2$