Question

solve for x [x+1/x+2]^2 = x+2/x+4​

Answer 1

Step-by-step explanation:

1/x+1 +2/x+2 = 4/x+4

(1(x+2)+2(x+1))/(x+1)(x+2) = 4/x+4

(x+2+2x+2)(x+4)=4(x+1)(x+2)

(3x+4)(x+4)=4(x^2+2x+x+2)

3x^2+12x+4x+16=4(x^2+3x+2)

3x^2+16x+16=4x^2+12x+8

x^2-4x-8=0

comparing with ax^2+bx+c=0

a=1 b= -4 c= -8

D=b^2-4ac=(-4)^2-4*1*(-8)

=16+32

=48>0

So x^2-4x-8=0 has two distinct real roots

they are (-b+root(b^2-4ac))/2a , (-b-root(b^2-4ac))/2a

=(-(-4)+root(48))/2(1), (-(-4)-root(48))/2(1))

=(4+root(16*3))/2 , (4-root(16*3))/2

=(4+4root3)/2, (4-4root3)/2

=2+2root3, 2-2root3