Question

The roots of the quadratic equation x^2-4x+3)+ x^2-6x+8)

Answer 1

${x}^{2} - 4x + 3 = 0 \\ {x}^{2} - 3x - x + 3 = 0 \\ x(x - 3) - (x - 3) = 0 \\ (x - 3)(x - 1) = 0 \\ therefore \: (x - 3) = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 3 \\ (x - 1) = 0 \\ x = 1 \\ second \: equation \: {x}^{2} - 6x + 8 = 0 \\ {x}^{2} - 4x - 2x + 8 = 0 \\ (x - 4)(x - 2) = 0 \\ therefore \: x = 4 \: and \: x = 2$