Question

28/(x^2-9)+x/(3-x)+7/(3+x)=1

Answer 1

Answer:

x=-2, or x=4

Step-by-step explanation:

Hello,

$(\forall x \in \mathbb{R}) \ x^2-9=x^2-3^2=(x-3)(x+3)$

So, assuming that we take x different from 3 and -3 (as dividing by 0 is not allowed)

$\dfrac{28}{x^2-9}+\dfrac{x}{3-x}+\dfrac{7}{3+x}=1\\ \\<=> 28+\dfrac{x(x+3)(x-3)}{3-x}+\dfrac{7(x+3)(x-3)}{3+x}=x^2-9\\\\<=>28-x(x+3)+7(x-3)=x^2-9\\ \\<=>28-x^2-3x+7x-21-x^2+9=0\\ \\<=>-2x^2+4x+16=0\\ \\<=>x^2-2x-8=0$

The sum of the zeroes is 2=4-2 and the product is -8=4*(-2), so we can conclude.

$x^2-2x-8=(x+2)(x-4)=0\\ \\<=> x = -2 \ or \ x = 4$

Thanks