Question

Explain Completing the Square method with the help of an example.

Answer 1

Answer is guven below :

Square method : This method i used to find the roots of a quadratic equation. We need to make the x or y variable polynomial of the form (ax + b)² and the other term be a constant part.

Method :

Let, any quadratic equation be
ax² + bx + c = 0,
where a, b and c are real numbers with non-zero a

Step 1 : In the first step, make the coefficient of x² = 1, by dividing both sides of the equation by a, where a must be non-zero.

So, we get
x² + (b/a)x + (c/a) = 0

Step 2 : In this step, we need to express apply identity get an expression of the form (a + b)² or, (a - b)² and a constant term.

Then,
x² + (b/a)x + (c/a) = 0

⇒ x² + {2 × x × b/(2a)} + (b/2a)² = (b/2a)² - c/a

⇒ (x + b/2a)² = (b/2a)² - c/a

⇒ (x + b/2a)² = b²/4a² - c/a

⇒ (x + b/2a)² = (b² - 4ac)/2a²

Thus, we have transformed the given equation as mentioned in Square method's definition.

Step 3 : In this step, we need to transform the (x + b/2a)² term into the (x + b/2a) term. Thus we need to take (1/2) power to both the sides.

Hence, we get

x + b/2a = ± {√(b² - 4ac)}/2a

Now, we take the constant part in Left Hand Side to the Right Hand Side.

⇒ x = - b/2a ± {√(b² - 4ac)}/2a

⇒ x = [- b ± √(b² - 4ac)]/2a

This be the solution of the given quadratic equation.

Example :

We take a quadratic equation :

x² - 12x + 35 = 0

⇒ x² - (2 × x × 6) + 6² = 6² - 35

⇒ (x - 6)² = 36 - 35,
since (a - b)² = a² - 2ab + b²

⇒ (x - 6)² = 1²

⇒ x - 6 = ± 1

⇒ x = 6 ± 1

Taking the positive sign, we get x = 6 + 1 = 7
and taking the negative sign, we get x = 6 - 1 = 5

Thus, the required solution be x = 6, 5

I hope it helps you.

Answer 2

Completing square method : In this method, the (ax² + bx + c=0) is converted in the form of [ (ax + b)² = -c + 2abx] , Where a is a positive number and not equal to 0.

=========================

Conversion :

××××××××××××××××××
We know, all equations are given in the form of

ax² + bx + c = 0
××××××××××××××××××

To make (ax + b)², there are few easy steps :-

(1) : Removal of 'a' from ax² [ divide whole equation]

$=> \frac{a {x}^{2} }{a} + \frac{bx}{a} + \frac{c}{a} = \frac{0}{a} \\ \\ => {x}^{2} + \frac{bx}{a} + \frac{c}{a} = 0 \\ \\ => {x}^{2} + \frac{bx}{a} = - \frac{c}{a}$

(2) : multiply and divide by 2 on (bx÷a)

${x}^{2} + \frac{2bx}{2a} = - \frac{c}{a}$

(3) : add [ b/2a]² on both sides,

${x}^{2} + 2( \frac{bx}{2a} ) + ( \frac{b}{2a} ) ^{2} = - \frac{c}{a} + (\frac{b}{2a} ) ^{2} \\ \\ \\ by \: formula \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ \\ {(x + \frac{b}{2a} )}^{2} = \frac{ - c}{a } + \frac{ {b}^{2} }{4 {a}^{2} } \\ \\ \\ x + \frac{b}{2a} = ( + - )\sqrt{ \frac{ - 4ac + {b}^{2} }{4 {a}^{2} } } \\ \\ x \: + \frac{b}{2a} = ( + - ) \frac{ \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \\ x = ( + - ) \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} - \frac{b}{2a} \\ \\ x \: = \frac{ - b( + - ) \sqrt{ {b}^{2} - 4ac } }{2a}$

×××××××××××××

Example :-

x² + 4x + 4 = 0

Solution :-

x² + 4x + 4 = 0

=> x² + 4x = -4
=> x² + 2(4x/2) = -4
=> x² + 2(4x/2) + (4/2)² = -4 +(4/2)²
=> (x + 4/2)² = -4 + 16/4
=> (x + 2)² = -4 + 4
=> x + 2 = 0
=> x = -2

I hope this will help you

(-: