Question

find the roots for the x-3/x=2(x is not equal to 0) quadratic equation by factorisation
I NEED THE PERFECT ANSWER​
PLEASE

Answer 1

Step-by-step explanation:

$\sf {x-{\dfrac{3}{x}}=2} \\ {\dashrightarrow}{\dfrac{{x}^{2}-3}{x}}=2 \\ {\dashrightarrow} {x}^{2}-3=2x \\ {\dashrightarrow} {x}^{2}-2x-3=0 \\ {\dashrightarrow} {x}^{2}-3x+x-3=0 \\ {\dashrightarrow} x (x-3)+1 (x-3)=0 \\ {\dashrightarrow} (x-3)(x+1)=0 \\ {\dashrightarrow} (x-3)=0 {\quad}or {\quad}(x+1)=0 \\ {\dashrightarrow} x=3 {\quad}or {\quad}x=(-1)$

$\therefore$${\underline{\boxed{\bf{x=(3,-1)}}}}$