Question

-2x^2+8x-8=0


a, 0;two real solution
b, 128; two real solution
c,0; 2 imaginary solution
d,0;one real solution

Answer 1

Hint: you could start by dividing both sides by -2.

Answer 2

divide both sides by -2 and will get , ... ???

can you make it sure ...!

Answer 3

are you here ?

Answer 4

yes

Answer 5

so than can you divide it ?

Answer 6

yes

Answer 7

and will get ---- ?

Answer 8

i don't if it is right or not but i got -8x-4=0

Answer 9

on simplifying you get -x^2+4x-4=0
-x^2+2x+2x_4=0
now factorize:
(-x+2)(-x+2)=0

Answer 10

so if you divide both sides by (-2) so than will result

-2x^2 +8x -8 =0 /(-2)

x^2 -4x +4=0

right ?

Answer 11

so what factorized will be eaxctly (x-2)^2

(x-2) squared

right ?

Answer 12

so than x-2=0
x=2

than will be right d. so one real solution

hope so much that is understandably sure right

i wait your message

bye

Answer 13

yes

Answer 14

yes @akkib89 there you have missed anything

ok ?

Answer 15

actually it will be mod(x-2)=0

Answer 16

-2x^2+8x-8=0

x^2 -4x +4 = 0

b^2-4ac

16 - 16 = 0

So..They have two equal Roots.....So..it means that....One SolN