Question

Na forma Resolutiva:
a) x²-7x+6=0
b)x²-x-12=0
c)x²+-3x-28=0
d)x²+12x+36=0
e)6x²-x-1=0

Answer 1

a) x²-7x+6=0

$x= \frac{ -b_{-}^+\sqrt{b^2- 4.a.c}}{2.a} = \frac{ -(-7)_{-}^+\sqrt{-7^2- 4.1.6}}{2.1}= \frac{ 7_{-}^+\sqrt{49- 24}}{2}= \frac{ 7_{-}^+\sqrt{25}}{2}= \frac{ 7_{-}^+{5}}{2} \\ \\ x^{1} = \frac{7+5}{2} = \frac{12}{2} =6 \\ \\ x^{2} = \frac{7-5}{2} = \frac{2}{2} = 1$

b)x²-x-12=0

$x= \frac{ -b_{-}^+\sqrt{b^2- 4.a.c}}{2.a} = \frac{ -(-1)_{-}^+\sqrt{-1^2- 4.1.(-12)}}{2.1}= \frac{ 1_{-}^+\sqrt{1+48}}{2}= \frac{ 1_{-}^+\sqrt{49}}{2}= \frac{ 1_{-}^+{49}}{2} \\ \\ x^{1} = \frac{1+49}{2} = \frac{50}{2} =25 \\ \\ x^{2} = \frac{49-1}{2} = \frac{48}{2} = 24$

c) x²+-3x-25=0 ( é + ou - 3x ? Responde nos comentários e eu faço ela aqui)

d)x²+12x+36=0

$x= \frac{ -12_{-}^+\sqrt{12^2- 4.1.36}}{2.1}= \frac{ -12_{-}^+\sqrt{144-144}}{2}= \frac{-12_{-}^+\sqrt{0}}{2}= \frac{ -12_{-}^+{0}}{2} \\ \\ x^{1} = \frac{-12+0}{2}= \frac{-12}{2}= -6 \\ \\ x^{2} = \frac{-12-0}{2} = \frac{-12}{2}=-6$

e)6x²-x-1=0

$x= \frac{ -1_{-}^+\sqrt{(-1)^2- 4.6.(-1)}}{2.6}= \frac{ -1_{-}^+\sqrt{1+24}}{12}= \frac{-1_{-}^+\sqrt{25}}{12}= \frac{ -1_{-}^+{5}}{12} \\ \\ x^{1} = \frac{-1+5}{12}= \frac{4}{12}= \frac{2}{6} \\ \\ x^{2} = \frac{-1-5}{12} = \frac{-6}{12}= \frac{-1}{4}$