Answer:
(x+ (b/2a))^ 2 = (b/2a) ^2 + c/a
Explanation:
ax²+bx=c
first we know the form of a square
(x+a)^2 = x^2 + 2ax + a^2
we have to write the given ax²+bx=c in the above form
for that we bring c on LHS
for the we subtract c from both sides
ax²+bx -c =c - c
=> ax²+bx - c = 0
now we see that in x^2 + 2ax + a^2 , term x^2 does not have any coefficient,
so in our equation we get rid of coefficient a from ax²+bx - c = 0
for that we divide both LHS and RHS by a
(ax²+bx - c)/a = 0/a
x²+bx/a - c/a = 0
now we have to write (b/a)x in form of 2ax
for we divide and multiply (b/a)x by 2
thus we have
x²+2 (b/2a)x - c/a = 0
now comparing above equation with x^2 + 2ax + a^2
we see that 2ax is same 2 (b/2a)x
thus (b/2a) is same as a in standard form of quadratic equation x^2 + 2ax + a^2
now we have to get a^2
as (b/2a) is same as a
we add and subtract square of (b/2a) on LHS
x²+2 (b/2a)x - c/a + (b/2a) ^2 - (b/2a) ^2 = 0
rearranging it
x²+2 (b/2a)x + (b/2a) ^2 - (b/2a) ^2 -c/a = 0
=> x^2 + (b/2a)x + (b/2a) ^2 -(b/2a) ^2 -c/a = 0
=> (x+ (b/2a))^ 2 - ((b/2a) ^2 + c/a ) = 0
=> (x+ (b/2a))^ 2 = (b/2a) ^2 + c/a answer