

To solve a quadratic equation of the form Ax^{2} + Bx + C = 0 using the method of completing the square
Do the following,
STEP[1] Substract the constant term from both sides of the quadratic equation, or just remove the constant term from the left hand side of equation ,change the sign of it
,if constant term is positive change it to negative if it is negative change it to positive and put it on the right hand side ,removing the original 0 from the right hand side .i.e Ax^{2} + Bx + C  C = 0  C
Ax^{2} + Bx + 0 = C
Ax^{2} + Bx = C
STEP[2] divide throughout by the numerical coefficient of the quadratic term or the coefficient of x^{2} ;
i.e , divide entire equation by 'A' the number in front of x^{2} .
now since 1x^{2} = 1 times x^{2} = x^{2}, when you see no number in front of
x^{2} know that 1 is really the number in front of x^{2} or that 1 is the
coefficient of x^{2},
further since 4/1 = 4 , y/1 = y etc.. more generally ,'any number divided by 1 gives the result of the number itself'. when you see no number in front of x^{2} ,you now know that 1 is the coefficient of x^{2} donot bother to divide by 1 ,just skip this step.
Ax^{2} + Bx = C
( Ax^{2}½A ) + Bx½A = C ½A
x^{2} + (B½A)x = C ½A
STEP[3]Now find half of the coefficient of x ,i.e find half of the number now in front of x ,to do that just multiply 1/2 to the coefficient of x ,since the coefficient of x is B/A ,we have ; (1/2)*B/A = B/2A
STEP[4]Now square half of the coefficient of x ,i.e square the result obtained in step[3] i.e is (B/2A)^{2}
STEP[5]Now add the square of half of the coefficient of x ,i.e
the result obtained in step[4] to both sides of equation obtained in step[2] i.e
x^{2} + (B½A)x + (B/2A)^{2} = (C ½A)+ (B/2A)^{2}
Now factorising the left hand side ,
since (B/2A)*(B/2A) = (B/2A)^{2}
and (B/2A)+(B/2A) = B/A
replacing (B/A)x with (B/2A)x+(B/2A)x in the equation immediately above
we have
x^{2} + (B/2A)x +(B/2A)x + (B/2A)^{2} = (C ½A)+ (B/2A)^{2}
putting the terms on the left hand side into groups of two
we have
[ x^{2} + (B/2A)x ] + [(B/2A)x + (B/2A)^{2} ] = (C ½A)+ (B/2A)^{2}
now factor each group on left hand side
x[x + (B/2A)] + (B/2A)[ x + (B/2A)] = (C ½A)+ (B/2A)^{2}
now factorise out x + (B/2A) from left hand side
[x + (B/2A)][x + (B/2A)] = (C ½A)+ (B/2A)^{2}
which gives :
[x + (B/2A)]^{2} =(C ½A)+ (B/2A)^{2}
Note : since x^{2} + (B½A)x + (B/2A)^{2} is a perfect square x^{2} + (B½A)x + (B/2A)^{2} = [x + (B/2A)] ^{2}
you donot have to factorise the Left hand side as I did ,by factorising I actually
proved why x^{2} + (B½A)x + (B/2A)^{2} = [x + (B/2A)] ^{2}
when you are using the method of completing the square skip the factorising bit.
Now to solve for x take the square root of both sides of equation
since the square root of [x + (B/2A)] ^{2} is [x + (B/2A)]
we have ;
[x + (B/2A)] = (+/) Ö [(C ½A)+ (B/2A)^{2}]
simplify right hand side and solve for x
Note , that in solving for x let [x + (B/2A)] = + Ö [(C ½A)+ (B/2A)^{2}]
and let [x + (B/2A)] =  Ö [(C ½A)+ (B/2A)^{2}]

Key to using steps 
Compare the quadratic equation which you have to solve (say, 2x^{2} + 3x + 1 = 0 )with the general form Ax^{2} + Bx + C = 0 ,
noting the values of A ,B and C (in this case A = 2 , B = 3 , C = 1 ) and substitute the
respective values of A , B , and C in each STEP above ,write each step (skip the factorising bit)...
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© 2004
Prakash Sukhu.








