Question

form a quadratic equation
1)  √x²+4 = (x²+1)²
2)x+1 + x+2  = 1
    x-1    x+3

Answer 1

ii) Given that
$\frac{x+1}{x-1} + \frac{x+2}{x+3} = 1$

Rewriting the fractions by taking LCM

$\frac{(x+1)(x+3)+ (x+2)(x-1)}{(x-1)(x+3)} = 1$

$\frac{x^2+4x+3+x^2+x-2}{x^2+2x-3} =1$

$\frac{2x^2+5x+1}{x^2+2x-3} =1$

$2x^2+5x+1 = x^2+2x-3$

$2x^2 - x^2 + 5x - 2x +1 + 3 = 0$

$x^2 + 3x +4 = 0$