Question

Solve the following quadratic equations by factorization:
(x-2/x-3)+(x-4/x-5)=10/3;x≠3,5

Answer 1

$\alpha=\frac{-7}{2} \\\beta=-6$

Step-by-step explanation:

$\frac{x-2}{x-3} +\frac{x-4}{x-5} =\frac{10}{3}\\\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)x-5)} =\frac{10}{3} \\\frac{x^2-5x-2x+10+x^2-3x-4x+12}{x^2-5x-3x+15} =\frac{10}{3}\\ cross multiply\\3(2x^2-14x+22)=10(x^2-8x+15)\\6x^2-42x+66=10x^2-80x+150\\6x^2-42x+66-10x^2+80x-150=0\\-4x^2+38x-84=0\\divide -ve\\4x^2-38x+84=0\\2(2x^2-19x+42)=0\\2x^2-19x+42=0$

a=2       b= -19     c=42

$D=b^2-4ac\\D=(-19)^2-4\times2\times42\\D=25$

$\alpha = \frac{-b\pm\sqrt{d} }{2a} \\\alpha=\frac{-b+\sqrt{d} }{2a} \\\alpha=\frac{-19+\sqrt{25} }{2\times2} \\\alpha=\frac{-19+5}{4} \\\alpha=\frac{-14}{4} \\\alpha=\frac{-7}{2}\\\beta =\frac{-b-\sqrt{d} }{2a} \\\beta =\frac{-19-\sqrt{25} }{2\times2} \\\beta =\frac{-19-5}{4} \\\beta =\frac{-24}{4} \\\beta=-6$