Question

Solve by completing the square method 2x2 x-4=0

Answer 1

Answer:

x^2+2x+4=0# can be written as

#ul(x^2+2xx x xx1 +1^2)-1^2+4=0#

and as #a^2+2ab+b^2=(a+b)^2#, this is

#(x+1)^2-1+4=0#

or #(x+1)^2+3=0#

Now, to solve the equation, we must convert it to form #a^2-b^2#, but as we have #+3#, we write it as #-(-3)#

and using imaginary numbers #-3=(isqrt3)^2#, as #i^2=-1# and #(sqrt3)^2=3#

Hence, we can write #(x+1)^2+3=0# as #(x+1)^2-(isqrt3)^2=0#

Using identity #a^2-b^2=(a+b)(a-b)#, this becomes

#(x+1+isqrt3)(x+1-isqrt3)=0#

i.e. either #x=-1-isqrt3# or #x=-1+isqrt3#.

Answer 2

⇒ 2x² + x = 4

Dividing both sides of the equation by 2, we get

⇒ x²+x/2 = 2

Now on adding (1/4)² to both sides of the equation, we get,

⇒ (x)² + 2 × x × 1/4 + (1/4)² = 2 + (1/4)²

⇒ (x + 1/4)² = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = ± √33-1/4

Therefore, either x = √33-1/4 or x = -√33-1/4