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The integers G and H are chosen such that
\[\frac{G}{x+5}+\frac{H}{x^2-4x}=\frac{x^2-3x+5}{x^3+x^2-20x}\]
for all real values of $x$ except -5, 0, and 4. Find $H/G$.

 Jun 22, 2022
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Simplify the left-hand side with a common denominator: \(\large{{(x^2-4x)g + (x+5)h \over x^3 + x^2 - 20x } = {x^2 - 3x + 5 \over x^3 + x^2 - 20x}}\)

 

Now, recall that the denominators of the fractions are equal, we can ignore them, giving us: \((x^2 - 4x)g + (x+5)h = x^2 - 3x + 5\)

 

The only way to get an \(x^2\) term on the left-hand side is if it comes from the term \((x^2 - 4x)g\), meaning \(gx^2= x^2\)

 

Thus, we know that \(g = 1\)

 

Likewise, doing the same thing with the \((x+5)h\) term, we notice that the coefficient of 5 can only come form the term 5h, meaning \(h = 1\)

 

Thus, the ratio \({ h \over g} = \color{brown}\boxed{1}\)

 Jun 22, 2022

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