Question

solve the following quadratic equation by factorisation and check the solutions
a(x^2+1) = x(a^2+1)

Answer 1

Given,

⇒ a( x² + 1 ) = x( a² + 1 )

⇒ ax² + a = a²x + x

⇒ ax² + a - a²x - x = 0

⇒ ax² - a²x - x + a = 0

Taking out ax common from first 2 terms, and -1 from last 2 terms,

⇒ ax( x - a ) - 1( x - a ) = 0

Taking out ( x - a ) common,

⇒ ( x - a ) ( ax - 1 ) = 0

⇒ ( x - a ) = 0 ÷ ( ax - 1 )

⇒ ( x - a ) = 0

•°• x = a

" Or "

⇒ ( x - a ) ( ax - 1 ) = 0

⇒ ( ax - 1 ) = 0 ÷ ( x - a )

⇒ ( ax - 1 ) = 0

⇒ ax = 1

•°• x = ( 1/a )

Hence, x = a or ( 1/a ).

Check :

When , x = a

⇒a( x² + 1 ) = x( a² + 1 )

Substitute the value of x = a,

⇒ a( a² + 1 ) = a( a² + 1 )

⇒ a³ + 1 = a³ + 1

When, x = ( 1/a ),

⇒ a( x² + 1 ) = x( a² + 1 )

Substitute the value of x = ( 1/a ),

⇒ a[ ( 1/a )² + 1 ] = ( 1/a ) ( a² + 1 )

⇒ a[ ( 1/a² ) + 1 ] = ( 1/a ) × a² + 1( 1/a )

⇒ a( 1/a² ) + a = ( a²/a ) + ( 1/a )

⇒ ( a/a² ) + a = a + ( 1/a )

⇒ ( 1/a ) + a = a + ( 1/a )

By commutative property of addition,

⇒ a + ( 1/a ) = a + ( 1/a )

Checked !!

The required answer is ( a ) and ( 1/a ).

Answer 2

a( x² + 1 ) = x( a² + 1 )

=> ax² + a = xa² + x

=> ax² - xa² + a - x = 0

=> ax( x - a ) - (- a + x ) = 0

=> ax( x - a ) - ( x - a ) = 0

=> ( x - a ) ( ax - 1 ) = 0

$=> x = a \: \: \: \: or \: \: \: \: x = \frac{1}{a}$




For checking,


Taking ( x = a ),

=> a( x² + 1 ) = x( a² + 1 )
=> a( a² + 1 ) = a( a² + 1 )
=> a³ + a = a³ + a
=> 0 = 0


Taking ( x = 1/a)

=> a( x² + 1 ) = x( a² + 1 )
=> a( 1/a² + 1 ) = 1/a ( a² + 1 )
=> a( 1 + a² )/a² = ( a² + 1 )/a
=> ( 1 + a² )/a = ( a² + 1 )/a
=> 0 = 0




Hence, checked that our solution is correct.