## Solving Quadratic Equations by Completing the Square Method:

Let us consider the quadratic equation by completing the square equation, x^{2} + 5x + 5 = 0. Here in the middle tem 5x cannot split into terms such that m + n = 5 and mn = 5. This means we cannot resolve the equation as product of two factors and therefore, it cannot be solved by factorisation method. This is the limitation of factorisation method of solving quadratic equation by factorisation. This method can be used only when it is possible to split the middle term and factorise the given quadratic polynomial.

Then, how to solve the quadratic equation where the quadratic polynomial cannot be factorised?

Let us take an example

**Example : Solve the quadratic equation 3x**^{2} – 5x + 2 = 0 by completing the square method

^{2}– 5x + 2 = 0 by completing the square method

Solution:

Given 3x^{2} – 5x + 2 = 0.

Here, the coefficient of x^{2} is 3 and it is not a perfect square.

Such quadratic equations are solved in two ways. Let us do both of them.

**(i) Multiply the equation throughout by 3.**

(3x^{2} – 5x + 2 = 0)x3

9x^{2} – 15x + 6 = 0

Now, half of the coefficient of x is ^{5}/_{2} .

b = ^{5}/_{2} and b^{2} = (^{5}/_{2})^{2}

So, 9x^{2} – 15x + (^{5}/_{2})^{2} – (^{5}/_{2})^{2} + 6 = 0

(3x)^{2} – 2(3x)(^{ 5}/_{2}) + (^{5}/_{2})^{2} = (^{5}/_{2})^{2} – 6

(3x)^{2} – 2(3x)(^{5}/_{2}) + (^{5}/_{2})^{2} = ^{25}/_{4} – 6

(3x – ^{5}/_{2})^{2} = ^{25 – 24}/_{4}

(3x – ^{5}/_{2}) = ± ^{1}/_{2}

3x = ± ^{1}/_{2} + ^{5}/_{2}

3x = ^{1}/_{2} + ^{5}/_{2} or 3x = – ^{1}/_{2} + ^{5}/_{2}

3x = ^{1+5}/_{2} or 3x = ^{-1+5}/_{2}

3x = ^{6}/_{2} or 3x = ^{4}/_{2}

x = ^{6}/_{2} x ^{1}/_{3} = 1 or x = ^{4}/_{2} x ^{1}/_{3}

x = 1 or x = ^{4}/_{6} = ^{2}/_{3}

Therefore, 1 and ^{2}/_{3} are the roots of the quadratic equation 3x^{2} – 5x + 2 = 0

** (ii) Dividing the equation throughout by 3.**

(3x^{2} – 5x + 2 = 0) ⁄3

Now, let us proceed as earlier

{b = ^{1}/_{2} x ^{5}/_{3} = ^{5}/_{6} ; b^{2} = (^{5}/_{6})^{2}}

So, x – ^{5}/_{3} x + (^{5}/_{6})^{2} – (^{5}/_{6})^{2} + ^{2}/_{3} = 0

(x – ^{5}/_{6})^{2} = (^{5}/_{6})^{2} – ^{2}/_{3}

(x – ^{5}/_{6})^{2} = ^{25}/_{36} – ^{2}/_{3}

(x – ^{5}/_{6})^{2} = ^{25-24}/_{36}

(x – ^{5}/_{6})^{2} = ^{1}/_{36}

(x – ^{5}/_{6}) = ± ^{1}/_{6}

x = ± ^{1}/_{6} + ^{5}/_{6}

x = +^{1}/_{6} +^{5}/_{6 }or x = –^{1}/_{6} +^{5}/_{6}

x = ^{6}/_{6} or x = ^{4}/_{6}

x = 1 or x = ^{2}/_{3}

Therefore, 1 and ^{2}/_{3} are the roots of the quadratic equation 3x^{2} – 5x + 2 = 0

## Quadratic Equations – Exercise 9.4 – Class 10

### 1. Solve the following quadratic equations by completing the square.

(i) 4x^{2} – 20x + 9 = 0

(ii) 4x^{2} + x – 5 = 0

(iii) 2x^{2} + 5x – 3 = 0

(iv) x^{2} + 16x – 9 = 0

(v)x^{2} – 3x + 1 = 0

(vi) t^{2} + 3t = 7

**Quadratic Equations – Exercise 9.4 – Solutions:**

**Solve the following quadratic equations by completing the square.**

**(i) 4x ^{2} – 20x + 9 = 0**

Solution:

4x^{2} – 20x + 9 = 0

4x^{2} – 18x – 2x + 9 = 0

2x(2x – 9) – 1(2x – 9) = 0

(2x – 1)(2x – 9) = 0

2x – 1 = 0 or 2x – 9 = 0

2x = 1 or 2x = 9

x = ^{1}/_{2} or x = ^{9}/_{2}

Therefore, ^{1}/_{2} and ^{9}/_{2} are the roots of the quadratic equation 4x^{2} – 20x + 9 = 0

** (ii) 4x ^{2} + x – 5 = 0**

Solution:

4x^{2} + x – 5 = 0

4x^{2} + 5x – 4x – 5 = 0

x(4x + 5) – 1(4x + 5) = 0

(4x + 5)(x – 1) = 0

4x + 5 = 0 or x – 1 = 0

4x = -5 or x = 1

x = –^{5}/_{4} or x = 1

Therefore, 1 and –^{5}/_{4} are the roots of the quadratic equation 4x^{2} + x – 5 = 0

** (iii) 2x ^{2} + 5x – 3 = 0**

Solution:

2x^{2} + 5x – 3 = 0

2x^{2} + 6x – x – 3 = 0

2x(x + 3) – 1(x + 3) = 0

(2x – 1)(x + 3) = 0

2x – 1 = 0 or x + 3 = 0

2x = 1 or x = -3

x = ^{1}/_{2} or x = -3

Therefore, x = ^{1}/_{2} or – 3 is the root of the Quadratic equation 2x^{2} + 5x – 3 = 0.

** (iv) x ^{2} + 16x – 9 = 0**

Solution:

x^{2} + 16x – 9 = 0

Half of the coefficient of x is ^{16}/_{2} . Therefore, b = 8 and b^{2} = (8)^{2} = 64

x^{2} + 16x + (^{16}/_{2})^{2} – (^{16}/_{2})^{2}– 9 = 0

x^{2} + 16x + (^{16}/_{2})^{2} = (^{16}/_{2})^{2} + 9

x^{2} + 2(8)(x) + 8^{2} = (^{16}/_{2})^{2} + 9

(x + 8)^{2} = 8^{2} + 9

(x + 8)^{2} = 64 + 9

(x + 8) = ±√73

x = – 8 ±√73

Therefore, x = – 8 ±√73 is the root of the Quadratic equation x^{2} + 16x – 9 = 0

**(v)x ^{2} – 3x + 1 = 0**

Solution:

x^{2} – 3x + 1 = 0

Half of the coefficient of x is ^{3}/_{2} . Therefore, b = ^{3}/_{2} and b^{2} = (^{3}/_{2})^{2}

x^{2} – 3x + (^{3}/_{2})^{2} – (^{3}/_{2})^{2} + 1 = 0

x^{2} – 2(x)(^{3}/_{2}) + (^{3}/_{2})^{2} = (^{3}/_{2})^{2} – 1

(x – ^{3}/_{2})^{2} = ^{9}/_{4} – 1

(x – ^{3}/_{2})^{2} = ^{9 – 4}/_{4}

(x – ^{3}/_{2})^{2} = ^{5}/_{4}

(x – ^{3}/_{2}) = ±√(^{5}/_{4})

x = ±√(^{5}/_{4}) + ^{3}/_{2}

x = ^{3}/_{2} ±^{√}^{5}/_{2}

x = ^{3}^{±√}^{5}/_{2}

Therefore, x = ^{3}^{±√}^{5}/_{2 }is the root of the quadratic equation x^{2} – 3x + 1 = 0

**(vi) t ^{2} + 3t = 7**

Solution:

t^{2} + 3t = 7

t^{2} + 3t – 7 = 0

Half of the coefficient of t is ^{3}/_{2} . Therefore, b = ^{3}/_{2} and b^{2} = (^{3}/_{2})^{2}

t^{2} + 3t + (^{3}/_{2})^{2} – (^{3}/_{2})^{2} – 7 = 0

t^{2} + 2(t)(^{ 3}/_{2}) + (^{3}/_{2})^{2} = (^{3}/_{2})^{2} + 7

(t + ^{3}/_{2})^{2} = ^{9}/_{4} + 7

(t + ^{3}/_{2})^{2} = ^{9+28}/_{4}

(t + ^{3}/_{2})^{2} = ^{37}/_{4}

(t + ^{3}/_{2}) = ±√(^{37}/_{4} )

(t + ^{3}/_{2}) = ±^{√}^{37}/_{2}

t = ±^{√}^{37}/_{2} – ^{3}/_{2}

t = ±^{√}^{37 – 3}/_{2}

or

t =^{-3 }^{± √}^{37}/_{2}

Therefore, t = ^{-3 }^{± √}^{37}/_{2} is the root of the quadratic equation t^{2} + 3t = 7

Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Quadratic equations – Exercise 9.3 – Class X