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I am trying to solve the quadratic equation $x^2-48x+432=0$ with out directly factoring OR using the quadratic equation. I am going with vieta. So $$r+s=48$$ and $$rs=432$$ I've already solved it by plugging in factors of $432$ the roots are $$r=12, s=36 $$

but that is too much guessing so I wanted to try a new approach.

Using the fact that $s=\frac{432}{r}$ and subbing in the sum gets me. $$r+\frac{432}{r}=48$$

Here lies my problem. Manipulating the equation to solve for $r$ gets me back to the original quadratic. What am I missing? Or is plugging the only way to do this?

Yiyuan Lee
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Chris
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  • Chris: should the formula be $r^2-48r+432 = 0$? – Warren Hill Feb 20 '14 at 12:51
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    @WarrenHill I believe it should be $x^2 - 48x + 432$ because $r,s$ seems to be used already for the roots. – Yiyuan Lee Feb 20 '14 at 12:52
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    Manipulating the equation to solve for r gets me back to the original quadratic. What am I missing? Nothing, you will always get back to the original. Your method works by trial and error (ie. finding factors and plugging) – Blah Feb 20 '14 at 12:52
  • @WarrenHill Yes it should but that is the circle I am stuck in. – Chris Feb 20 '14 at 12:53

3 Answers3

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Complete the square $$x^2-48x+432=0\iff x^2-2\cdot x\cdot24+24^2=576-432$$

$$\iff(x-24)^2=(12)^2$$

lab bhattacharjee
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  • Hmm he didn't want to use the quadratic equation, which is equivalent to completing the square. – user21820 Feb 20 '14 at 12:49
  • Vieta's formulas seem to be using knowledge of how solutions to a quadratic behave, which you get from the quadratic formula as well. – R R Mar 27 '14 at 23:58
  • @RR: No, Vieta's formula follow directly from the factorization of the quadratic and comparing of coefficients, and works even when the underlying field has characteristic $2$, whereas the quadratic formula doesn't work in that case! (And you should use "@"+ otherwise only the author of the answer will be notified of your comment.) – user21820 May 21 '14 at 12:57
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$(r-s)^2 = (r+s)^2 - 4rs$

So you can get $(r-s)$ and hence $r = \frac{(r+s)+(r-s)}{2}$

user21820
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  • Note that you will in general get two possible solutions for $(r-s)$, one for each possible solution for $r$. – user21820 Feb 20 '14 at 12:51
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Given a quadratic equation $x^2-bx+c=0$, then solving $r+s=b$ and $rs=c$ is equivalent to finding $r,s$ such $(x-r)(x-s)=x^2-bx+c$. So, Vieta's formulas are nice but do not simplify anything.

lhf
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  • So I can use the quadratic to solve a product and sum simultaneously. But I can't use a product and sum to solve the quadratic. – Chris Feb 20 '14 at 13:07
  • @Chris: Doesn't my answer use the product and sum to obtain the solution to the quadratic as you wanted? In general there are 'similar tricks' to do the same for cubic equations and quartic equations, but you'll need to learn Galois theory to understand what is really behind those tricks, otherwise all it would seem to you is coincidence. – user21820 May 21 '14 at 13:03