Question

1)encontre as raizes das equacoes incompletas do 2º  grau abaixo.
a)x²-49=0
b)x²=1
c)2x²-50=0
d)3(x²-1)=24
e)5(x²-1)=4(x²+1)
f)(x-3)(x+4)+8=x
 

Answer 1

$x = \frac{-b \pm \sqrt{-b^2 -4*a*c}}{2*a}$

A)
$x^{2} - 49 = 0$

a=1, b=0, c=−49

Δ=b2−4ac
Δ=(0)2−4*(1)*(−49)
Δ=0+196
Δ=196

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{0 \pm \sqrt{196}}{2*1} \\ \\ x = \frac{0 \pm 14}{2} \\ \\ x' = \frac{0 + 14}{2} \\ \\ x' = \frac{14}{2} \\ \\ x' = 7 \\ \\ x'' = \frac{0 - 14}{2} \\ \\ x'' = \frac{-14}{2} \\ \\ X'' = -7$

S = {7, - 7}

B)
$x^{2} = 1 \\ \\ x^{2} -1 = 0$

a=1, b=0, c=−1
Δ=b2−4ac
Δ=(0)2−4*(1)*(−1)
Δ=0+4
Δ=4

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{0 \pm \sqrt{4}}{2*1} \\ \\ x = \frac{0 +{2}}{2} \\ \\ x' = \frac{{2}}{2} \\ \\ x' = 1 \\ \\ \\ x'' = \frac{0 -{2}}{2} \\ \\ x'' = \frac{-{2}}{2} \\ \\ x'' = -1$

S= {1, -1}

C)
$2x^{2} -50 = 0$

a = 2, b = 0, c = -50
Δ=b2−4ac
Δ=(0)2−4*(2)*(−50)
Δ=0+400
Δ=400

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{0 \pm \sqrt{400}}{2*2} \\ \\ x = \frac{0 \pm 20}{2*2} \\ \\ x' = \frac{20}{4} \\ \\ x' = \frac{20}{4} \\ \\ x' = 5 \\ \\ \\ x'' = \frac{0 - 20}{4} \\ \\ x'' = \frac{-20}{4} \\ \\ x'' = \frac{20}{4} \\ \\ x'' =- 5$

S = {5, -5}

D)

$3( x^{2} -1) = 25 \\ \\ 3x^2 - 3 - 25 = 0$

a=3, b=−3, c=−25
Δ=b2−4ac
Δ=(−3)2−4*(3)*(−25)
Δ=9+300
Δ=309

Não existe raiz exata para 309, então fica deste modo: 

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-3) \pm \sqrt{309}}{2*3} \\ \\ x = \frac{3 \pm \sqrt{309}}{6} \\ \\ x' = \frac{3 + \sqrt{309}}{6} \\ \\ x'' = \frac{3-\sqrt{309}}{6}$

S = {$\frac{3 + \sqrt{309}}{6}, \frac{3 - \sqrt{309}}{6}$}

E)

$5( x^{2} -1) = 4( x^{2} +1) \\ \\ 5 x^{2} -5 = 4x^{2} + 4 \\ \\ 5 x^{2} -5 - 4x^2 -4 = 0 \\ \\ x^{2} -9 = 0$

a = 1, b = 0, c = -9

Δ=b2−4ac
Δ=(0)2−4*(1)*(−9)
Δ=0+36
Δ=36

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-0 \pm \sqrt{36}}{2*1} \\ \\ x = \frac{0 \pm 6}{2} \\ \\ x = \frac{0+ 6}{2} \\ \\ x = \frac{6}{2} \\ \\ x' = 3 \\ \\ \\ x'' = \frac{0 - 6}{2} \\ \\ x'' = \frac{ - 6}{2} \\ \\ x'' = - 3$

S = {3, -3}

F)

$(x - 3) (x + 4) + 8 = x \\ \\ x^2 + 4x - 3x - 12 + 8 = x \\ \\ x^2 + x - 4 = x \\ \\ x^2 + x - 4 -x = 0 \\ \\ x^2 - 4 = 0$

a = 1, b= 0, c = - 4 
Δ=b2−4ac
Δ=(0)2−4*(1)*(−4)
Δ=0+16
Δ=16

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-0 \pm \sqrt{16}}{2*1} \\ \\ x = \frac{0 \pm 4}{2} \\ \\ x' = \frac{ 4}{2} \\ \\ x ' = 2 \\ \\ \\ x = \frac{0 - 4}{2} \\ \\ x'' = \frac{- 4}{2} \\ \\ x ' = -2$

S= {2, -2}