## How to solve a quadratic equation by completing the square :-

### Today in this article I want to discuss a special topic how to solve a quadratic equation by completing the square,which is a very important topic for the students of standard(x). To start my explanation I want to express the idea to learn,

## ** What is a quadratic equation ?

### Answer:- An equation which contains only one variable and the degree of variable is 2 is known as a quadratic equation. e.g; x² - x - 6 = 0 is a quadratic equation of one variable x. Quadratic equations are of two types :-

### (i) Pure quadratic equations :-Examples,

(a) 2x² - 18 = 0

(b) y² - 64 = 0

(ii) Impure or mixed quadratic equations:-

Examples, (a) 3x² - 4x -7 =0

(b) y² + 6y + 5 = 0

## ***what is standard quadratic equation?

### Answer:- A quadratic equation in the form of ax² + bx + c = 0 where, a never equal to zero is known as standard quadratic equation, where a,b,and c are the coefficient of x², x and constant term respectively.

Now we will solve the standard quadratic equation, ax² + bx + c =0

### Solution:- 1st Method :-

ax² + bx + c =0

=>ax² + bx = - c

=>4a(ax² + bx) = - 4ac [∵ coefficient of x² term is a so its four time i.e; 4× a = 4a has been multiplied in both sides ]

=>4a²x² + 4abx = - 4ac

=>(2ax)² +2.2ax.b +(b)² = b² - 4ac [∵ a² - 2ab + b² = ( a - b)²]

=>(2ax + b)² = b² - 4ac

=>(2ax + b) = ± √b² - 4ac

=> 2ax = - b ± √b² - 4ac

=> x = (- b ± √b² - 4ac)/2a, which is known as the quadratic formula of quadratic equation.

also

b² - 4ac is known as discriminant of the quadratic equation and the nature of the roots of any quadratic equation always depends on the discriminant.

2nd Method :-

ax² + bx + c =0

=> ax² + bx = - c

=> ax²/a + bx/a = - c/a [ dividing both sides by a]

=> x² + bx/a = - c/a

=> (x)² + 2.x. b/2a + ( b/2a)² = (b/2a)² - c/a

=>(x + b/2a)² = b² /4a² - c/a

=>(x + b/2a)² = (b² - 4 ac)/4a²

=> (x + b/2a) = ± √(b² - 4 ac)/2a

=> x = - b/2a ± √(b² - 4 ac)/2a

=> x = [-b ± √(b² - 4 ac)]/2a Solved.

A quadratic equation can be solved in three different methods and they are:-

## (i) Factorisation Method:-

### This method is totally based on middle term factors method and sometimes square formula can be used.

For the illustration of this method ,Let us take a quadratic equation 5x² - x -6 = 0, now we will solve it using factorisation method:-

Solve the given equations by factorisation method:-

### Question (1) 5x² - x -6 = 0

### =>5x² - (6 -5)x - 6 = 0

=> 5x² + 5x - 6x - 6 = 0

=> 5x(x + 1) - 6(x + 1) = 0

=> ( x + 1) ( 5x - 6) = 0

Either :- or :-

(x + 1) = 0 5x - 6 = 0

=> x = -1 => 5x = 6

=> x = 6/5

∴ The roots of the given roots = ( -1 or 6/5) solved.

## Rule (1):-

### (a)1st we will find out the coefficient of x-term here coefficient of x -term is -1

(b) we will find the sign before constant term, here constant term is 6 and sign before it is negative.

(c) we will multiply coefficient of x² term (here it is 5) and coefficient of x term(here it is 6) and we will get = 6 × 5 = 30

(d) we need to find two factors of 30 such that when they are subtracted we will get 1

(e) Here two factors of 30 will be 6 and 5.

∵ negative sign is there before constant term, so we will make = 6 - 5= 1

(f)Now 6 - 5 will be multiplied by x keeping correct sign of the products.

(g) In the 3rd line of the sum we will get 4 terms and from 1st two terms we have taken 5x as a common factor ,then we get (x+1)similarly taking -6 we will also get (x+1)

(h) In the 4th line when common factor (x+1) is taken out we will get (5x-6) another fector.Now we have got (x+1)(5x-6) = 0

(i) ∵ The product of any two positive number can never be zero unless their individual value is zero.

(j) We have taken (x+1) = 0 so, x = - 1

similarly when (5x -6) = 0 there will be x

=6/5

∴ The roots of the quadratic equation is = ( -1 or 6/5)

## Question(2):- 4x² + 4x + 1 = 0

### Solution :-

### 4x² + 4x + 1 = 0

=> (2x)² + 2.2x.1 + (1)² = 0

=> ( 2x + 1)² = 0

=> ( 2x + 1)( 2x +1) = 0

∴ Either:- or:-

2x + 1 = 0 2x + 1 = 0

=>2x = -1 => 2x = - 1

=> x = - ½ => x = - ½

∴ The roots of the quadratic equation is = ( -½ and -½ )

## Rule (2):-

### (a) In 1st line we will break all terms of L.H.S in the form of a² + 2ab + b²

(b)In 2nd line They are written as in the form ( a +b)² = 0

Here ( a +b)² = ( 2x + 1)²

(c) In 3rd line ( 2x + 1)² have been written 2 times as a product.

(d) In 4th line we have written (2x + 1) = 0

and we have got x = - ½

similarly putting (2x + 1) = 0 we have got x = - ½

∴ The roots are = ( -½ and -½ )

## (ii) Quadratic formula Method:-

### In this method we only can solve a quadratic equation by Quadratic formula. Let , we do a sum in this method.

Question(3):- Solve the given equation by Quadratic formula:-

y² - 2y - 3 = 0

Here

a = 1

b = - 2

and c = - 3

∴ y = (- b ± √b² - 4ac)/2a,

= [ - ( -2) ± √(-2)² - 4.1.(-3)]/ 2.1

= (2 ± √ 4 + 12) / 2

= (2 ± √ 16) / 2

= ( 2 ± 4)/2

∴ Either :- or:-

y = (2 + 4)/2 y = (2 - 4)/2

= 6/2 = - 2/2

= 3 = -1

∴ Roots of the equation = ( 3 or - 1) Solved.

## Rule (3) :-

### (a) 1st we will write the values of coefficient of y² term (a),coefficient of y term(b) and constant term/ term without y(c)

(b) Now we will write y which will be equal to our quadratic formula.

(c) Putting all the values of a,b and c we have got y = (2 + 4)/2 and y = ( 2 - 4) /2

Now calculating we have got two values of y and they are ( 3 or - 1)

## (iii) Completing the square or completing square method:-

### Our main discussion is how to solve a quadratic equation by completing the square so we will now focuss on our main topic.

## Solve the following quadratic equations using completing square :-

## **** Question(4)

### Solution:-

### 1st Method:-

### 2x² - x -3 = 0

=> 2x² - x = 3

=>8(2x² - x) = 3 × 8 [∵ coefficient of x² term is 2 so its four time i.e; 4× 2 = 8 has been multiplied in both sides ]

=>16x² - 8x = 24

=> (4x)² - 2.4x.1 + (1)² = 24 + 1 [adding 1 in both sides.]

=> ( 4x -1 )² = 25 [∵ a² - 2ab + b² = ( a - b)²]

=> 4x - 1 = + 5 or -5[taking square root in both sides]

∴

Either:- Or:-

4x -1 = 5 4x - 1 = - 5

=> 4x = 5 +1 => 4x = - 5 +1

=> x = 6/4 => x = - 4/4

=> x = 3/2 => x = - 1

∴ Roots of the given equation is = (3/2,or -1) Solved.

2nd Method :-

2x² - x -3 = 0

=> 2x² - x = 3

=>2x²/2 - x/2 = 3/2 [dividing both sides by 2]

=> x² - x/2 = 3/2

=>(x)² - 2.x.¼ + (¼ )² = 3/2 + (¼ )²

=>(x - ¼)² = 3/2 + 1/16[∵ a² - 2ab + b² = ( a - b)²]

=>(x- ¼ )² = (3 × 8 + 1 × 1)/16

=>(x- ¼ )² = 25/16

=>(x -¼) = ± 5/4 [taking square root in both sides]

∴ Either:- Or:-

(x -¼) = 5/4 (x -¼) = - 5/4

=> x = 5/4 + ¼ =>x = - 5/4 +¼

=> x = (5×1+1×1)/4 =>x = (-5×1+1×1)/4

=>x = 6/4 =>x = (- 5+1)/4

=>x =3/2 =>x = - 4/4

=>x = -1

∴ Roots of the given equation is = ( 3/2,-1)solved.

## Rule (4):-

### For 1st Method:-

### (i) We will rewrite the equation in the form ax² + bx = c

(ii) We will add to both sides the term needed to complete the square.

(iii) Now we will fector the perfect square trinomial.

(iv) Lastly we will solve the resulting equation by using the square root property.

Maintaining the above rules in the above question we have found, roots of the above equation are = ( 3/2 and -1)

## Rule(5):-

### For 2nd Method :-

### (i) We will rewrite the equation in the form ax² + bx = c

(ii) Both sides will be divided by 2

(iiii) We will add to both sides the term needed to complete the square.

(iv) Now we will fector the perfect square trinomial.

### (v) Now we will solve the resulting equation by using the square root property.

vi) Taking square root in both sides we have got two values of (x -¼ ) as 5/4 and -5/4

(vii) 1st case, we have taken (x -¼) = 5/4 and simplifying R.H.S terms we have got x = 3/2 and in 2nd case, we have taken (x -¼) = - 5/4 and simplifying R.H.S terms we have got x = - 1

∴ The roots of the given equation is = (3/2 and -1)

## ***Question (5) 3x² +10x + 3 = 0

### Solution:-

### 1st Method:-

### 3x² +10x + 3 = 0

=>3x² +10x = - 3

=> 12( 3x² +10x) = -3 × 12 [∵coefficient of x² term is 3 so its four time i.e; 4× 3 = 12 has been multiplied in both sides ]

=>36x² + 120x = -36

=>(6x)² + 2.6x.10 + (10)² = (10)² - 36 [adding 10² in both sides.]

=> (6x +10)² = 100 - 36 [∵ a² - 2ab + b² = (a - b)²]

=> (6x +10)² = 64

=> 6x +10 = + 8 or -8 [ taking square root in both sides]

∴ Either:- or:-

6x +10 = 8 6x - 10 = - 8

=> 6x = 8 -10 =>6x = - 8 -10

=> x = -2/6 =>x = - 18/6

=> x = -⅓ =>x = - 3

∴ The roots of the given equation = (- ⅓ ,or -3)Solved.

### 2nd Method:-

### 3x² +10x + 3 = 0

=>3x² +10x = - 3

=>3x²/3 +10x/3 = - 3/3

=> x² + 10x/3 = -1[dividing both sides by 3]

=>(x)² + 2.x.5/3 +(5/3)² = (5/3)² -1 [ adding (5/3)² in both sides ]

=>(x + 5/3)² = 25/9 -1 [∵ a² - 2ab + b² = ( a - b)²]

=>(x + 5/3)² = (25 -9)/9

=>(x + 5/3)² = 16/9

=>(x + 5/3) = + 4/3 or -4/3

Either:- or:-

x + 5/3 = 4/3 x + 5/3 = - 4/3

=> x = - 5/3 + 4/3 =>x = - 5/3 - 4/3

=> x = (- 5 + 4)/3 =>x = (- 5 - 4)/3

=> x = - ⅓ =>x = -9/3

=>x = - 3

∴ The roots of the given equation is = (-⅓ or -3) Solved.

## Rule(6):-

### For 1st Method:-

### (i) We will rewrite the equation in the form ax² + bx = c

(ii) We will add to both sides the term needed to complete the square.

(iii) Now we will fector the perfect square trinomial.

(iv) Lastly we will solve the resulting equation by using the square root property.

Following the above rules in above equation,

(v) 1st case,we have taken (6x+10) = 8 and we have got x = -⅓ and in 2nd case,we have taken (6x+10) = - 8 and we have got x = - 3

∴ Roots of the equation = (- ⅓ or - 3)

## Rule(7):-

### For second method:-

### (i) We will rewrite the equation in the form ax² + bx = c

(ii) Both sides will be divided by 3

(iii) Now we will add both sides the term needed to complete the square.

(iv)We will fector the perfect square trinomial.

(v) Then we will solve the resulting equation by using the square root property.

### vi)Taking square root in both sides we have got two values of (x +5/3 ) as 4/3 and -4/3

## FREQUENTLY ASKED QUESTIONS:-
## What is standard Quadratic equation?

Answer:- A quadratic equation in the form of ax² + bx + c = 0 where, a never equal to zero is known as standard quadratic equation,where a,b,and c are the coefficient of x², x and constant term respectively.

## Three examples of quadratic equation.

Answer :- Three examples of quadratic equation are :-
(i) x² - x - 12 = 0
(ii) 2x² +5 x + 3 = 0
(iii) 3y² - y - 4 = 0

## what is discriminant of a quadratic equation?

Answer :- b² - 4ac in the quadratic formula is known as discriminant of a quadratic equation.

## Under what condition the roots of a quadratic equation become equal?

Answer:- In a quadratic equation
if b² - 4ac = 0, then both roots of that equation will be equal to each-other.

##
Under what condition the standard quadratic 'ax² + bx + c = 0' become a linear equation?

Answer :- If a = 0 then the standard quadratic 'ax² + bx + c = 0' become a linear equation.

## What is standard Quadratic equation?

Answer:- A quadratic equation in the form of ax² + bx + c = 0 where, a never equal to zero is known as standard quadratic equation,where a,b,and c are the coefficient of x², x and constant term respectively.

## Three examples of quadratic equation.

Answer :- Three examples of quadratic equation are :- (i) x² - x - 12 = 0 (ii) 2x² +5 x + 3 = 0 (iii) 3y² - y - 4 = 0

## what is discriminant of a quadratic equation?

Answer :- b² - 4ac in the quadratic formula is known as discriminant of a quadratic equation.

## Under what condition the roots of a quadratic equation become equal?

Answer:- In a quadratic equation if b² - 4ac = 0, then both roots of that equation will be equal to each-other.

## Under what condition the standard quadratic 'ax² + bx + c = 0' become a linear equation?

Answer :- If a = 0 then the standard quadratic 'ax² + bx + c = 0' become a linear equation.