Question

Resolva a equação 
a) x²+9x-442=0
b)n² +32n =0
c) 7z²-112=0

por favor me ajudem 

Answer 1

a)
x²+9x-442=0
Δ=81+1768=1849
√Δ=43

$x'= \frac{-9+43}{2} = \frac{34}{2} =17$

$x"= \frac{-9-43}{2} = \frac{-52}{2} =-26$

S={-26,17}

b)
n²+32n=0

n(n+32)=0
n'=0

n+32=0
n"=-32

S={-32,0}

c)
7z²-112=0
7z²=112
z²=112÷7
z²=16
z=+ ou-√16
z'=4
z"=-4

s+{-4,+4]

Answer 2

a) x²+9x-442=0

A= 1

B= 9

C= - 442

∆= b² - 4ac

∆= 9² - 4 • 1 • ( - 442)

∆= 81 + 1768

∆= 1849

x= - b ± √∆ /2a

x= - 9 ± √1849 /2 • 1

x'= - 9 + 43 /2 = 34/2 = 17

x''= - 9 - 43/2 = - 52/2 = - 26

S= ( - 26 , 17)

b)n² +32n =0

n • (n + 32) = 0

n= 0

n + 32 = 0

n= - 32

S= ( 0 , - 32)

c) 7z²-112=0

7z² = 112

z²= 112/7

z²= 16

z= ± √16

z= ± 4

S= ( 4 , - 4)