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# Roots of the polynomial by the complete square method

What is a complete square- The complete square is the term which can be written as a square of another term, as an example: 4 = 2², 9 = 3², 16 = 4², 25 = 5², etc.

What is a complete square quadratic Polynomial– The complete square quadratic polynomial is the algebraic expression which can be written into the form of a square of another algebraic expression, as an example: 4x² + 10x + 25 = (2x + 5)², 25x² ─20x + 4 = (5x ─2)². Factorizing a polynomial by the method of the complete square means, first of all adding and subtracting a term to the given polynomial such that it becomes a complete square that eases the factors of the polynomial.

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What is the term to be added to get the complete square- As an example, we have a quadratic equation 4x² + 10x ─ 24

Comparing this expression with the standard form of square quadratic a² +b² + 2ab = (a + b)²

In the given expression 4x² + 10x ─ 24, arranging the term 10x  equivalent to 2ab, 4x² equivalent to a², 4x² = (2x)² which means we have got the term ‘a’, now with the help of the value of 2ab and a which are 10x and 2x respectively, we can get the value of the term ‘b’ as follows.

STEP-1

a² + 2ab + b² = 4x² + 10x ─ 24

a² + 2ab + b² = (2x)² +2× 2x ×b – 24

2× 2x ×b = 10x

b = 10x/4x = 5/2

STEP 2 Adding and subtracting the value of b² in the LHS of the equation as follow

(2x)² +10x -24 + (5/2)² -(5/2)²

Rearranging the terms representing the complete square algebraic expression a² +2ab +b²

(2x)² +10x + (5/2)² -24 -(5/2)²

=$\left ( 2x +\frac{5}{2} \right )^{^{2}}-24-\frac{25}{4}$

=$\left ( 2x +\frac{5}{2} \right )^{^{2}}-\frac{121}{4}$

Write the above expression in the form of (a² -b²)

=$\left ( 2x +\frac{5}{2} \right )^{^{2}}-\left ( \frac{11}{2} \right )^{2}$

=$\left ( 2x +\frac{5}{2}+\frac{11}{2} \right )\left ( 2x+\frac{5}{2}-\frac{11}{2} \right )$

$\left ( 2x+ \frac{16}{2} \right )\left ( 2x-\frac{6}{2} \right )$

=(2x + 8)(2x─3)

STEP-3

= (2x + 8)(2x─3)

Which gives

2x + 8 = 0, x = – 4

2x─3 = 0,

$x = \frac{3}{2}$

Hence the required roots of the polynomial are  -4 and 3/2

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Conclusion: Fundamental rule of making an algebraic quadratic polynomial a square quadratic is adding and subtracting the term   $\left ( \frac{b}{2a} \right )^{2}$in the given expression.

Other Examples: Solve the following equation by the complete square method.

(a) 2x²–5x – 3 =0   (b)3x² +7x + 2=0

(a) 2x²–5x – 3 =0

Solution.

2x²–5x – 3 =0

First of all multiplying the equation by 2, so that the term (2x²) could become a complete square term

4x² – 10 x – 6 = 0

Comparing with complete square

a² – 2ab + b² = 0

(2x)² – 2 × 2x × b + b² = 0

2 × 2x × b = 10x

b = 10/4

b = 5/2

Adding and subtracting b² = (5/2)² =25/4 in the LHS of the quadratic equation

(2x)² – 10x + (5/2)² – 25/4 – 6 = 0

(2x – 5/2)² – 49/4

(2x – 5/2)² – (7/2)²

(2x – 5/2 + 7/2) (2x – 5/2 – 7/2)

1/2 × 1/2 (4x – 5 + 7) (4x – 5 – 7) = 0

(4x + 2) (4x – 12) = 0

2 × 4 (2x +1) (x – 3) = 0

(2x – 1) (x – 3) = 0

x = 1/2, x = 3

Hence the required roots are 1/2 and 3

(b)3x² +7x + 2=0

Solution.

Multiplying the given quadratic equation by 3 so that the quadratic term (3x²) becomes a complete square

9x² +21x + 6=0

Comparing with complete square

a² +2ab + b² = 0

(3x)² +2 × 3x × b + b² = 0

2 × 3x × b = 21x

b = 21/6 = 7/2

9x² +21x +( 7/2)²– 49/4 +6=0

The expression in bold is complete square

(3x)² +21x + (7/2)² -25/4 = 0

(3x + 7/2)² – (5/2)² = 0

(3x + 7/2 + 5/2)(3x +7/2 -5/2) = 0

(3x + 6)(3x + 1) = 0

x =-2, x = -1/3

Hence the required roots of the quadratic roots are -2 and -1/3

Solution:

(a) 2x²–5x – 3 =0

First of all, making the quadratic term ‘2x²’ a complete square by multiplying whole of the equation by 2.

We shall have

4x²–10x – 6 =0

(2x)² – 10x –6 =0

Adding and subtracting the following term in the equation, in the equation 4x²–10x – 6 =0 ,b =-10, a= 4

$\left ( \frac{b}{2a} \right )^{2}=\left ( \frac{-10}{4} \right )^{2}=\frac{100}{16}=\frac{25}{4}$

Adding and subtracting (25/4) in the LHS of the quadratic equation,(2x)² – 10x –6 =0

$\left ( 2x \right )^{2}-10x- 6 + \frac{25}{4}- \frac{25}{4}$

$\left (2x-\frac{5}{2} \right )^{2}-\frac{49}{4}$

$\left (2x-\frac{5}{2} \right )^{2}- \left ( \frac{7}{2} \right )^{2}$
$\left ( 2x-\frac{5}{2}+\frac{7}{2} \right )\left ( 2x-\frac{5}{2}-\frac{7}{2} \right )$
$\left [ 2x-\frac{\left ( 5-7 \right )}{2} \right ]\left [ 2x-\frac{\left ( 5+7 \right )}{2} \right ]$
(2x +1)(2x –6) =0
Which gives,
$x = \frac{-1}{2}, 3$
Solution:
(b) 3x² +7x + 2=0
Making the term 3x², a complete square by multiplying the equation by 3.
9x² +21x + 6=0
(3x)² +21x + 6 =0
Adding and subtracting the equation by the following term.
$\left ( \frac{b}{2a} \right )^{2} = \left ( \frac{21}{6} \right )^{2}=\frac{49}{4}$
$\left ( 3x \right )^{2}+21x +6 +\frac{49}{4}-\frac{49}{4}$
$\left ( 3x \right )^{2}+21x +\left ( \frac{7}{2} \right )^{2} +6-\frac{49}{4}$
$\left ( 3x \right )^{2}+21x +\left ( \frac{7}{2} \right )^{2} -\frac{25}{4}$
$\left ( 3x+ \frac{7}{2} \right )^{2}- \left ( \frac{5}{2} \right )^{2}$
$\left ( 3x+ \frac{7}{2}+\frac{5}{2} \right )\left ( 3x+\frac{7}{2}-\frac{5}{2} \right )$
$\left ( 3x+6 \right ) \left ( 3x+1 \right )$=0
$x = -2, \frac{-1}{3}$

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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