QuadraticEquationsSolutions

Quadratic Equations - Solutions by Completing the Square Method

Consider the following identities

x2+2ax+a2=(x+a)2

x2-2ax+a2=(x-a)2


These trinomials are perfect squares. The highest power x2 has unity (i.e., 1) as its coefficient. In this case, the term without an x (the constant term) must equal the square of half the coefficient of x. Therefore, if the terms x2 and x are given, the square may be completed by adding the square of half the coefficient of x.

If ax2+bx+c=0

Transposing ax2+bx=-c

Dividing by a to make the coefficient of x2 = 1 we get

To complete the square, we add the square of half the coefficient of x, or

   =         , to both sides of the equation


We obtain      


Taking the square roots on both sides

which is, as you know, the formula.


Example 1

Solve using the completing the square method

x2+14x=32

a=1, b=14, c= -32


(x + 7)2 = 32 + 49

(x + 7)2 = 81

x + 7 = ± 9

i.e., x = 9 - 7 or x = -9 - 7

x = 2 or x = -16


Example 2

Solve the equation x2-2x=8 by completing the square,

comparing x2 - 2x = 8 with x2+ b / a x = c / -a

a = 1, b = -2, c = -8


               x2 - 2x + (1)2 = 8 + (1)2

               (x - 1)2 = 8 + 1

               (x - 1)2 = 9

Taking square roots on both the sides

               x2 - 1 = ± 3

x-1=3 or x-1=-3

x=1+3 or x=-3+1

x=4 or x=-2


Example 3

Solve by completing the square

       3x2-5x= 2

Taking the square roots on both sides

Try these questions:

Solve by completing the square

  1. (2x-3) (3x+4) = 0

  2. 3x2+7x = 6

  3. 4x2-2x = 3

  4. x2 = 0.6x - 0.05

  5. 5(x+2) = 4(2x-1)(x+1)-16

  6. 2x2-7x = 4

  7. 12x2-cx-20c2 = 0

Answers to Practice Problems

Solutions: To solve by completing the square.

  1. (2x-3) (3x+4) = 0


    On expanding we get

    2x(3x+4) -3 (3x+4) = 0

           6x2 +8x -9x -12 = 0

              6x2-x-12 = 0

    ⇒        6x2-x = 12


    Dividing throughout by 6 we get

    x2 - x / 6= 12/6

    x2 - x / 6 = 2        (1)

    Comparing with x2 b / -a x = c / -a we get

    a=6 b= -1, c= -2


  2. To solve 3x2 + 7x = 6 by completing the square

    Dividing throughout by 3 we get

    ∴ x = 2/3, -3 are the roots of the equation.


  3. To solve 4x2-2x = 3 by completing the square

    Dividing throughout by a = 4 we get

    x2 2 / -4 x = 3/4

    Comparing with x2+ b/a x= c/-a

    a=4 b=-2, c=-3

    Taking square roots on both the sides we get


  4. To solve by completing the square

           x2 = 0.6x - 0.05

    Transposing

           x2-0.6x = -0.05

    Comparing with x2+b / a x = c / -a

    a = 1, b = - 0.6, c = - 0.05


    Adding (b / 2a)2 = (-0.6/2)2 = (-0.3)2 = .09 to both sides we get

    x2-0.6x+0.09 = -0.05+0.09

    x2-0.6x+(0.3)2 = 0.04

           (x-0.3)2 = (0.2)2

    Taking square roots on both the sides

    x - 0.3 = ± 0.2

    ⇒ x-0.3 = 0.2 or x - 0.3 = - 0.2

    ⇒        x = 0.2 + 0.3 or x = - 0.2 + 0.3

    ⇒        x = 0.5 or x = 0.1

    Roots of the equations are x = 0.1, 0.5


  5. To solve 5(x+2) = 4(2x-1)(x+1) -16 by completing the square

    We first obtain the form ax2 + bx + c = 0

    5x+10 = 4[(2x -1)x+(2x - 1) * 1] -16

    5x+10 = 4[2x2-x+2x-1]-16

    5x+10 = 4[2x2 + x - 1] -16

    5x+10 = 8x2 + 4x - 4 - 16

    ⇒ 8x2 + 4x - 20 - 5x - 10 = 0

    ⇒        8x2 - x - 30 = 0

    ⇒        8x2 - x = 30

    Dividing throughout by 8


  6. To solve 2x2-7x = 4 by completing the square

    Dividing throughout by a = 2


  7. To solve 12x2-cx-20c2 = 0 by completing the square Transposing and dividing by a = 12 we get

    x2c / -12 x = 20 / 12 c2

    Comparing with x + (b/-a) x = -c/a

    a = 12, b = -c, c = - 20c2


        to both sides

    Taking square roots on both sides

    Roots of the equations are