## SOLVING QUADRATIC EQUATIONS USING COMPLETING THE SQUARE METHOD

Solving Quadratic Equations Using Completing the Square Method :

In this section, we will learn how to solve quadratic equations using completing the square method.

## Steps Involved in Completing the Square Method

Step 1 :

In the given quadratic equation ax2 + bx + c = 0, divide the complete equation by a (coefficient of x2).

If the coefficient of x2 is 1 (a = 1), the above process is not required.

Step 2 :

Move the number term (constant) to the right side of the equation.

Step 3 :

In the result of step 2, write the "x" term as a multiple of 2.

Examples :

6x should be written as 2(3)(x).

5x should be written as 2(x)(5/2).

Step 4 :

The result of step 3 will be in the form of

x2 + 2(x)y  =  k

Step 4 :

Now add y2 to each side to complete the square on the left side of the equation.

Then,

x2 + 2(x)y + y2  =  k + y2

Step 5 :

In the result of step 4, if we use the algebraic identity

(a + b)2  =  a2 + 2ab + b2

on the left side of the equation, we get

(x + y)2  =  k + y2

Step 6 :

Solve (x + y)2  =  k + yfor x by taking square root on both sides.

## Solving Quadratic Equations Using Completing the Square Method - Examples

Example 1 :

Solve by completing the square method

x2 + 6 x – 7 = 0

Solution :

x2 + 6x – 7  =  0

x2 + 6x  =  7

x2 + 2 (x) (3)  =  7

x2 + 2 (x) (3) + 32  =  7 + 32

(x + 3)2  =  7 + 9

(x + 3)2  =  16

x + 3  =  ±√16

x + 3  =  ±4

 x + 3  =  4x  =  4 - 3x  =  1 x + 3  =  -4x  =  -4 - 3x  =  -7

Hence the solution is { -7, 1 }.

Example 2 :

Solve by completing the square method

x2 + 3x + 1  =  0

Solution :

x2 + 3x + 1  =  0

x2 + 3x  =  -1

x2 + 2⋅x⋅(3/2)  =  -1 + (3/2)2

x2 + 2x⋅(3/2) + (3/2) =  -1 + (3/2)2

[x + (3/2)] =  -1 + (9/4)

[x + (3/2)] =  5/4

[x + (3/2)]  =  √(5/4)

[x + (3/2)]  =  ±5/2

 [x + (3/2)]  =  √5/2x  =  (√5/2) - (3/2)x  =  (√5 - 3)/2 [x + (3/2)]  =  -√5/2x  =  (-√5/2) - (3/2)x  =  (-√5 - 3)/2

Hence the solution is { (-√5 - 3)/2 and (√5 - 3)/2 }.

Example 3 :

Solve by completing the square method

2x2 + 5x - 3  =  0

Solution :

2x2 + 5x - 3  =  0

2x2 + 5x  =  3/2

Divide the equation by 2.

x² + (5/2)x  =  (3/2)

Half of coefficient of x is 5/4.

By adding (5/4)2 on both sides, we get

x² + (5/2)x + (5/4)2  =  (3/2) + (5/4)2

x² + (5/2)x + (5/4)2  =  (3/2) + (5/4)

[ x + (5/4) ]2  =  (3/2) + (25/16)

[ x + (5/4) ]2  =  49/16

[ x + (5/4) ]2  =  √(49/16)

x + (5/4) = ± 7/4

 x + (5/4)  =  7/4x  =  7/4 - (5/4)x  =  (7 - 5)/4x  =  2/4x  =  1/2 x + (5/4)  =  -7/4x  =  -7/4 - (5/4)x  =  (-7 - 5)/4x  =  -12/4x  =  -3

Hence the solution is {-3, 1/2}.

Example 4 :

Solve by completing the square method

4x2 + 4bx – (a2 - b2)  =  0

Solution :

4x2 + 4bx – (a2 - b2)  =  0

4x2 + 4bx  =  (a2 - b2)

x2 + bx  =  (a2 - b2)/4

Half of coefficient of x is b/2. By adding (b/2)2 on both sides, we get

x2 + bx + (b/2)2  =  (a2 - b2)/4 + (b/2)2

[ x + (b/2) ]2  =  (a2 - b2)/4 + (b2/4)

[ x + (b/2) ]2  =  a2/4

[ x + (b/2) ]  =  √(a2/4)

[x + (b/2)]  =  ±(a/2)

 x + (b/2)  =  (a/2)x =  (a/2) - (b/2)x  =  (a - b)/2 x + (b/2)  =  (-a/2)x =  (-a/2) - (b/2)x  =  -(a + b)/2 After having gone through the stuff given above, we hope that the students would have understood how to solve quadratic equations by completing the square method.

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