Question

please answer this question
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Answer 1

$ax^2+a=a^2x+x\\\\ax^2+a-a^2x-x=0\\\\ax^2-a^2x+a-x=0\\\\ax(x-a)+(a-x)=0\\\\-ax(a-x)+(a-x)=0\\\\(a-x)(1-ax)=0\\\\\implies a-x=0\;\;or\;\;1-ax=0\\\\\implies \boxed{x=a\;\;or\;\;x=\dfrac{1}{a}}$

Answer 2

Solution

Method I :

ax² + a = a²x + x

⇒ ax² + a = x(a² + 1)

⇒ ax² + a = (a² + 1)x

⇒ ax² - (a² + 1)x + a = 0

Comparing with ax² + bx + c = 0 we get,

• a = a
• b = - (a² + 1)
• c = a

Discriminant D = b² - 4ac

= { - (a² + 1) }² - 4(a)(a)

= (a² + 1)² - 4a²

= (a²)² + 2(a²)(1) + 1² - 4a²

= (a²)² + 2a² + 1 - 4a²

= (a²)² - 2a² + 1

= (a²)² - 2(a²)(1) + 1²

= (a² - 1)²

[ Because x² - 2xy + y² = (x - y)² ]

Using quadratic formula

x = ( - b ± √D )/2a

Substituting the values

⇒ x = [ - { - (a² + 1) } ± √(a² - 1)² ]/2(a)

⇒ x = {a² + 1 ± (a² - 1)} / 2a

⇒ x = {a² + 1 + (a² - 1)} / 2a or x = {a² + 1 - (a² - 1)} / 2a

⇒ x = (a² + 1 + a² - 1)/2a or x = (a² + 1 - a² + 1)/2a

⇒ x = 2a²/2a or x = 2/2a

⇒ x = a or x = 1/a

Method II :

ax² + a = a²x + x

⇒ ax² + a = a²x + x

⇒ ax² - a²x - x +a = 0

⇒ ax(x - a) - 1(x - a) = 0

⇒ (ax - 1)(x - a) = 0

⇒ ax - 1 = 0 or x - a = 0

⇒ ax = 1 or x = a

⇒ x = 1/a or x = a

Hence, a and 1/a are the roots of the equation.