# Solve the quadratic equations by formula 1 `a(x^2+1) = (a^2+1)x` , a is not 0 2 `4x^2 - 4ax+(a^2-b^2)=0`

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(1) `a(x^2+1)=(a^2+1)x`

`ax^2+a=(a^2+1)x`

`ax^2-(a^2+1)x+a=0` Use the quadratic formula:

If `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)`

Here `a=a,b=-(a^2+1),c=a` so

`x=(a^2+1+-sqrt((a^2+1)^2-4(a)(a)))/(2a)`

`=(a^2+1+-sqrt(a^4+2a^2+1-4a^2))/(2a)`

`=(a^2+1+-sqrt(a^4-2a^2+1))/(2a)`

`=(a^2+1+-sqrt((a^2-1)^2))/(2a)`

`=((a^2+1)+-(a^2-1))/(2a)`

so `x=(a^2+1+a^2-1)/(2a)=(2a^2)/(2a)=a`

or

`x=(a^2+1-a^2+1)/(2a)=2/(2a)=1/a`

(2) `4x^2-4ax+(a^2-b^2)=0`

Here a=4,b=-4a and `c=a^2+b^2`

`x=(4a+-sqrt((-4a)^2-4(4)(a^2-b^2)))/8`

`=(4a+-sqrt(16a^2-16a^2+16b^2))/8`

`=(4a+-sqrt(16b^2))/8`

`=(4a+-4b)/8`

so `x=1/2(a+b)` or `1/2(a-b)`