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Solving Quadratic Equations using Completing the Square Method

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Algebra is a mathematical science that studies diverse symbols that represent quantities that do not have a set value or amount associated with them but instead vary or change over time in connection to some other feature. In algebra, such symbols are known as variables, and the numbers associated with them are known as coefficients. They can be expressed in several ways, including English alphabets. In other terms, algebra is the study of how numbers are represented using letters or symbols without highlighting their true meanings.

Quadratic Equation

A quadratic equation is an algebraic statement of the second degree in x. The quadratic equation is written as ax2 + bx + c = 0, with a and b being the coefficients, x being the variable, and c being the constant factor. The coefficient of x2 must not be zero (a ≠ 0) for an equation to be classified as a quadratic equation. The x2 term comes first, followed by the x term, and finally the constant term when writing a quadratic equation in standard form. Integral values, rather than fractions or decimals, are typically used to denote the numeric values a, b, and c.

Completing the Square Method

A method or approach for converting a quadratic polynomial or an equation into a perfect square with an additional constant is called completing the Square Method. Using the formula or approach of the complete square, the quadratic equation in the variable x, ax2 + bx + c, where a, b and c are the real values ​​except a = 0, can be transformed or converted to a perfect square with an additional constant.

The completing the square formula is given by,

ax2 + bx + c ⇒ a(x + m)2 + n

Where, 

m = b/2a,

n = c – (b2/4a)

Here, m can be any real number and n is a constant.

Derivation:

Suppose the given quadratic equation is ax2 + bx + c. It can be written as,

ax2 + bx + c = a (x2 + (b/a)x) + c

= a (x2 + 2 (bx/2a) + (b/2a)2 – (b/2a)2) + c

= a (x2 + 2 (bx/2a) + (b/2a)2 – b2/4a2) + c

= a (x2 + 2 (bx/2a) + (b/2a)2) + (c – b2/4a)

Using the property (a + b)2 = a2 + b2 + 2ab we get,

= a (x + b/2a)2 + (c – b2/4a)

Replacing b/2a by a real number m and (c – b2/4a) by a constant value we get,

= a(x + m)2 + n

This derives the formula for completing the square method.

Sample problems

Question 1: Use completing the square method to solve: x2 + 4x – 21 = 0.

Solution:

We have, a = 1, b = 4 and c = –21.

Find the value of m and n.

m = 4/2 = 2

n = –21 – (16/4) = –21 – 4 = –25

So, the equation is solved as,

(x + 2)2 – 25 = 0

x + 2 = ±5

x = 3, –7

Question 2: Use completing the square method to solve: x2 + 10x + 21 = 0.

Solution:

We have, a = 1, b = 10 and c = 21.

Find the value of m and n.

m = 10/2 = 5

n = 21 – (100/4) = 21 – 25 = –4

So, the equation is solved as,

(x + 5)2 – 4 = 0

x + 5 = ±2

x = –3, –7

Question 3: Use completing the square method to solve: x2 + 6x – 27 = 0.

Solution:

We have, a = 1, b = 6 and c = –27.

Find the value of m and n.

m = 6/2 = 3

n = –27 – (36/4) = –27 – 9 = –36

So, the equation is solved as,

(x + 3)2 – 36 = 0

x + 3 = ±6

x = 3, –9

Question 4: Use completing the square method to solve: x2 + 12x – 13 = 0.

Solution:

We have, a = 1, b = 12 and c = –13.

Find the value of m and n.

m = 12/2 = 6

n = –13 – (144/4) = –13 – 36 = –49

So, the equation is solved as,

(x + 6)2 – 49 = 0

x + 6 = ±7

x = 1, –13

Question 5: Use completing the square method to solve: x2 + 20x + 19 = 0.

Solution:

We have, a = 1, b = 20 and c = 19.

Find the value of m and n.

m = 20/2 = 10

n = 19 – (400/4) = 19 – 100 = –81

So, the equation is solved as,

(x + 10)2 – 81 = 0

x + 10 = ±9

x = –1, –19

Question 6: Use completing the square method to solve: x2 + 6x – 16 = 0.

Solution:

We have, a = 1, b = 6 and c = –16.

Find the value of m and n.

m = 6/2 = 3

n = –16 – (36/4) = –16 – 9 = –25

So, the equation is solved as,

(x + 3)2 – 25 = 0

x + 3 = ±5

x = 2, –8

Question 7: Use completing the square method to solve: x2 – 4x – 12 = 0.

We have, a = 1, b = –4 and c = –12.

Find the value of m and n.

m = –4/2 = –2

n = –12 – (16/4) = –12 – 4 = –16

So, the equation is solved as, 

(x – 2)2 – 16 = 0

x – 2 = ±4

x = 6, –2



Last Updated : 20 Mar, 2024
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