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The question:

Find values of $a,b,c.$ if $\displaystyle \frac{x^2+1}{x^2+3x+2} = \frac{a}{x+2}+\frac{bx+c}{x+1}$

My working so far:

http://i.imgur.com/VegifVa.jpg

How do I isolate $a$, $b$ and $c$?

juantheron
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Jaskirat Boparai
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2 Answers2

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For the right-hand side, let's obtain a common denominator. To that end, we obtain

$$\frac{a}{x+2}+\frac{bx+c}{x+1}=\frac{a(x+1)+(bx+c)(x+2)}{x^2+3x+1}=\frac{bx^2+(a+2b+c)x+(a+2c)}{x^2+3x+1}$$

Equating this last expression to $\frac{x^2+1}{x^2+3x+1}$ we see that

$$b=1$$

$$a+2b+c=0$$

and

$$a+2c=1$$

Can you finish from here?

SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION

We see that since $b=1$, the problem is reduced to solving a linear system of two equations for the two remaining unknowns, $a$ and $c$. We have $$a+c=-2$$and$$a+2c=1$$Upon solving, we find $a=-5$ and $c=3$.

Mark Viola
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  • Thanks, this helped a lot :) I was able to do the question without the spoiler, i think understand the concept now. – Jaskirat Boparai Aug 29 '15 at 07:00
  • That's great! I am pleased to hear. Just le t me know whenever you need help with other problems. – Mark Viola Aug 29 '15 at 07:02
  • Well I do have one question, and since I cant ask a new one, since the site wont let me, Ill ask here: The question is "Find the values of k for which x^2 + kx +k +3 = 0 has real roots. My answers were k=6 or k=-2, but im not sure if these are right, since questions like these have inequalities. Can you help? – Jaskirat Boparai Aug 29 '15 at 07:24
  • Sure. Find $k$ so that $k^2-4(k+3)\ge 0$. Rewriting, $(k-6)(k+2)\ge 0$. This, $k\ge 6$ or $k\le -2$. – Mark Viola Aug 29 '15 at 15:19
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Given $$\displaystyle \frac{x^2+1}{x^2+3x+2} = \frac{a}{x+2}+\frac{bx+c}{x+1} = \frac{a(x+1)+(bx+c)(x+2)}{x^2+3x+2}$$

So $$x^2+1 = bx^2+(a+2b+c)x+(a+2c)$$

Now equating Coefficients, we get $b=1$ and $(a+2b+c) = 0$ and $a+2c=1$

So Put $b=1$ in $a+2b+c=0\Rightarrow a+c=-2$ and above $a+2c=1$

so we get $c=3$ and $a=-5$

So we get $(a,b,c) = (-5,1,3)$

juantheron
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