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
=
=
Write the above expression in the form of (a² -b²)
=
=
–
=(2x + 8)(2x─3)
STEP-3
= (2x + 8)(2x─3)
Which gives
2x + 8 = 0, x = – 4
2x─3 = 0,
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 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
Adding and subtracting (25/4) in the LHS of the quadratic equation,(2x)² – 10x –6 =0
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