What should be added to \(\cfrac{1}{x^2-7x+12}\) to get \(\cfrac{2}

1.	What should be added to \(\cfrac{1}{x^2-7x+12}\) to get \(\cfrac{2}{x^2-6x+8}\)A) \(\cfrac{1}{x^2-5x+6}\)B) \(\cfrac{-1}{x^2-5x-6}\)C) \(\cfrac{4}{(x+3)(x+2)}\)D) \(\cfrac{2}{(x+3)(x-2)}\)
Answer» Correct option is (A) \(\frac{1}{x^2-5x+6}\) Let p(x) should be added to \(\frac{1}{x^2 -7x+12}\) to get \(\frac{2}{x^2 -6x+8}\) \(\therefore\) \(p(x)+\frac{1}{x^2 -7x+12}\) \(=\frac{2}{x^2 -6x+8}\) \(\Rightarrow\) \(p(x)=\frac{2}{x^2 -6x+8}\) \(-\frac{1}{x^2 -7x+12}\) \(=\frac{2}{(x-4)(x-2)}-\frac{1}{(x-4)(x-3)}\) \(=\frac{2(x-3)-(x-2)}{(x-2)(x-3)(x-4)}\) \(=\frac{2x-6-x+2}{(x-2)(x-3)(x-4)}\) \(=\frac{x-4}{(x-2)(x-3)(x-4)}\) \(=\frac{1}{(x-2)(x-3)}\) \(=\frac{1}{x^2-5x+6}\) Correct option is A) \(\cfrac{1}{x^2-5x+6}\)

What should be added to \(\cfrac{1}{x^2-7x+12}\) to get \(\cfrac{2}{x^2-6x+8}\)A) \(\cfrac{1}{x^2-5x+6}\)B) \(\cfrac{-1}{x^2-5x-6}\)C) \(\cfrac{4}{(x+3)(x+2)}\)D) \(\cfrac{2}{(x+3)(x-2)}\)

Answer»

Correct option is (A) \(\frac{1}{x^2-5x+6}\)

Let p(x) should be added to \(\frac{1}{x^2 -7x+12}\) to get \(\frac{2}{x^2 -6x+8}\)

\(\therefore\) \(p(x)+\frac{1}{x^2 -7x+12}\) \(=\frac{2}{x^2 -6x+8}\)

\(\Rightarrow\) \(p(x)=\frac{2}{x^2 -6x+8}\) \(-\frac{1}{x^2 -7x+12}\)

\(=\frac{2}{(x-4)(x-2)}-\frac{1}{(x-4)(x-3)}\)

\(=\frac{2(x-3)-(x-2)}{(x-2)(x-3)(x-4)}\) \(=\frac{2x-6-x+2}{(x-2)(x-3)(x-4)}\)

\(=\frac{x-4}{(x-2)(x-3)(x-4)}\)

\(=\frac{1}{(x-2)(x-3)}\) \(=\frac{1}{x^2-5x+6}\)

Correct option is A) \(\cfrac{1}{x^2-5x+6}\)