-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

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

parthkohli · Answer

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

jhonyy9 · Answer

divide both sides by -2 and will get , ... ???

can you make it sure ...!

jhonyy9 · Answer

are you here ?

this · Answer

yes

jhonyy9 · Answer

so than can you divide it ?

this · Answer

yes

jhonyy9 · Answer

and will get ---- ?

this · Answer

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

anonymous · Answer

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

jhonyy9 · Answer

so if you divide both sides by (-2) so than will result 

-2x^2 +8x -8 =0  /(-2)

x^2 -4x +4=0  

right ?

jhonyy9 · Answer

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

jhonyy9 · Answer

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

anonymous · Answer

yes

jhonyy9 · Answer

yes @akkib89 there you have missed anything 

ok ?

anonymous · Answer

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

anonymous · Answer

-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