Question

2. Solve the following quadratic equations using quadratic formula

Answer 1

a(x² + 1) = x(a² + 1)

ax² + a = a²x + x

ax² - (a² + 1)x + a = 0

where, a = a , b = -(a² + 1) , c = a

D = b² - 4ac

D = [-(a² + 1)]² - 4(a)(a)

D = a⁴ + 1 + 2a² - 4a²

D = a⁴ + 1 - 2a²

√D = √(a² - 1)²

√D = (a² - 1)

x = $\large{\bold{\frac{-b +_-\sqrt{D}}{2a}}}$

x = $\large{\bold{\frac{a^2 + 1 +_-(a^2 - 1)}{2a}}}$

taking (+ve).

x = $\large{\bold{\frac{a^2 + 1 + a^2 - 1}{2a}}}$

x = $\large{\bold{a}}$

taking (-ve).

x = $\large{\bold{\frac{a^2 + 1 - a^2 + 1}{2a}}}$

x = $\large{\bold{\frac{2}{2a}}}$

x = $\large{\bold{\frac{1}{a}}}$

Answer 2

Given Quadratic equation is

a( x² + 1 ) = x( a² + 1 )

=>ax² + a - ( a² + 1 )x = 0

=> ax² - ( a² + 1 )x + a = 0 Compare this

with Ax² + Bx + C = 0 we get ,

A = a , B = -( a² + 1 ) , C = a ,

Discreminant ( D ) = B² - 4AC

= [ -( a² + 1 ) ]² - 4 × a × a

= ( a² + 1 )² - 4a²

= ( a² )² + 2a² + 1 - 4a²

= ( a² )² - 2a² + 1

D = ( a² - 1 )²

By Quadratic Formula :

x = [ -B± √D ]/2A

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

= [ a² + 1 ± ( a² - 1 ) /2a

Now ,

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

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