Question

1. Solve the following quadratic equations
by completing the square method.
(a) x²-6x +4=0​

Answer 1

Given, Quadratic equation i.e x² - 6x + 4 = 0

• To find, value of x

Solution :

• Quadratic equation = x² - 6x + 4

→ x² - 6x + 4 = 0

→ x² - 6x = - 4

• (a - b)² = a² + b² - 2ab
• Add and subtract 3

→ (x)² - 2*x*3 + (3)² - (3)² = - 4

→ (x - 3)² - 9 = - 4

→ (x - 3)² = 9 - 4

→ (x - 3)² = 5

→ x - 3 = ± √5

• Either

→ x - 3 = √5

→ x = 3 + √5

• Or

→ x - 3 = - √5

→ x = 3 - √5

•°• The values of x are 3 + √5 and 3 - √5

Answer 2

Given Equation:-

• x² - 6x + 4 = 0

To Find:-

• Roots of the equation

Method Used:-

• Method of Completing the Square

Solution:-

Solving the Equation,

$\sf :\implies\:x^2 - 6x + 4 = 0$

Transporting 4 to Right side,

$\sf :\implies\:x^2 - 6x = -4$

Now to make it a identity, Adding 9 to both the side,

$\sf :\implies\:x^2 - 6x +9 = -4 + 9$

$\sf :\implies\:x^2 - 2\times 3\times x +9 = 5$

It is clearly seen that it is forming the identity i.e. $\bf a^2 - 2ab +b = (a +b)^2$

$\sf :\implies\:(x-3)^2= 5$

$\sf :\implies\:x-3= ±\sqrt{5}$

$\bf :\implies\:x= 3+\sqrt{5}\:and\:3-\sqrt{5}$

Hence, The Roots of the given Equation are 3+√5 and 3-√5.

━━━━━━━━━━━━━━━━━━━━━━━━━