(A) x^2 + 6x + 9 = 0

(B) 8x^2+ 5x - 6 = 0

(C) (x + 4)^2 - 36 = 0

Query (A)

Query (B)

Query (C)

solution given:

(A) x^2 + 6x + 9 = 0

doing middle term

x^2+  (3+3)x+9=0

x^2 +3x+3x +9=0

taking common from two each term.

x(x+3)+3(x+3)=0

(x+3)(x+3)=0

either

x+3=0

x=-3

(B) 8x^2+ 5x - 6 = 0

Comparing above equation with

ax^2+bx+c=0,

we get,

a=8

b=5

c=-6

now

we have

now substituting value:

taking positive

taking negative

(C) (x + 4)^2 - 36 = 0

(x + 4)^2 - 6^2 = 0

it is in the form of x²+y²:(x+y)(x-y)

so (x + 4)^2 - 6^2 can be written as (x+4+6)(x+4-6)

above equation becomes

(x+10)(x-2)=0

either

x=-10

or

x=2

Step-by-step explanation: