The correct equations in the question are:
(a) [tex]\dfrac{x}{x^2+x-2}[/tex] (b) [tex]\dfrac{6-x}{x^2-3x-54}[/tex]
Answer:
Step-by-step explanation:
From above:
a) [tex]\dfrac{x}{x^2+x-2}[/tex]
We are to find the roots of the quadratic equation at the denominator.
i.e.
x²+x-2 = x² + 2x - x - 2
= x(x+2) - 1(x+2)
= (x -1 ) (x + 2)
Thus;
[tex]\mathbf{\dfrac{x}{x^2+x-2} = \dfrac{A}{x-1} + \dfrac{B}{x+2}}[/tex]
b) [tex]\dfrac{6-x}{x^2-3x-54}[/tex]
The quadratic equation roots can be calculated by finding two numbers that if we multiply then, we will get -54 and if we add or subtract them, we will get -3
x²-3x -54 = x² - 9x +6x - 54
= x(x-9) + 6(x-9)
= (x + 6) (x - 9)
[tex]\mathbf{\dfrac{6-x}{x^2-3x-54} = \dfrac{A}{x+6} + \dfrac{B}{x-9}}[/tex]
