How to solve completing the square method questions
Answers
Answered by
3
Steps are following :-
1.) Transfer the constant term in R.HS.
2.) Divide the whole equation by the coficient of x^2
3.) Add the square of half of the coficient of x in both sides.
Now you will get perfect square..
For Example :-
2x^2 + 10x + 12 = 0
=> 2x^2 +10x = - 12
=> x^2 + 5x = - 6
=> x^2 + 5x + (5/2)^2 = - 6+(5/2)^2
=> (x +5/2)^2 = - 6 + 25/4
=> (x+5/2)^2 = 1/4
=> x + 5/2 = +- 1/2
=> x = 5/2 + - 1/2
x = 2 and 3
1.) Transfer the constant term in R.HS.
2.) Divide the whole equation by the coficient of x^2
3.) Add the square of half of the coficient of x in both sides.
Now you will get perfect square..
For Example :-
2x^2 + 10x + 12 = 0
=> 2x^2 +10x = - 12
=> x^2 + 5x = - 6
=> x^2 + 5x + (5/2)^2 = - 6+(5/2)^2
=> (x +5/2)^2 = - 6 + 25/4
=> (x+5/2)^2 = 1/4
=> x + 5/2 = +- 1/2
=> x = 5/2 + - 1/2
x = 2 and 3
Answered by
1
It's simple
take any Quadratic Equation such as: x^2 - 4x +4 = 0
we have, x^2 - 4x +4 =0
so, x^2 -4x = -4
we have to show it in the form of a whole square
so, (x)^2 -2(2)(x) + (2)^2 = -4 + (2)^2
becoz (a-b)^2 = a^2 + b^2 - 2ab
where we took a = x and b = 2
so, (x-2)^2 = 0
x-2 = 0
so, x = 2
take any Quadratic Equation such as: x^2 - 4x +4 = 0
we have, x^2 - 4x +4 =0
so, x^2 -4x = -4
we have to show it in the form of a whole square
so, (x)^2 -2(2)(x) + (2)^2 = -4 + (2)^2
becoz (a-b)^2 = a^2 + b^2 - 2ab
where we took a = x and b = 2
so, (x-2)^2 = 0
x-2 = 0
so, x = 2
Similar questions