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So i'm doing some problems from quadratic equations, this one in particular is supposed to be solved with vieta's formulas, I'm quite lost at what to do since there is a minus here, for example in solving $x_1^2+x_2^2$ for the same equation, its just a matter of adding $+2x_1x_2 -2x_1x_2$ to it and then inserting vieta's formulas in, but I have no idea what to do with -.

\begin{aligned} & x^2-3 x-10=0 \\ & x_1^2-x_2^2=? \end{aligned}

CiaPan
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Rahz
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  • This is hard to follow. Are $x_1,x_2$ the roots of the original quadratic? How do you distinguish between them or are you only trying to compute this up to sign? And, of course, we have $x_i^2=3x_i+10$ which simplifies the problem (or, of course, you could just solve the original quadratic). – lulu May 23 '23 at 11:57
  • Can you compute both $x_1-x_2$ and $x_1+x_2$ – Aditya Dwivedi May 23 '23 at 11:57
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    Unless you know $x_1>x_2$ or $x_2<x_1$, this can only be determined up to a sign. – daruma May 23 '23 at 11:58
  • X1 and X2 are roots, solutions of the original quadratic equation, the result is supposed to be 21 and -21 , but I have no idea how to get to there. Here's the full text of the problem: X1 and X2 are the solutions of the quadratic equation, without solving the equation find x1^2-x2^2. – Rahz May 23 '23 at 12:05

2 Answers2

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HINT…try using $$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$$

David Quinn
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The two zeroes $ \ x_1 \ $ and $ \ x_2 \ $ imply that $ \ x_1^2 - 3 x_1 - 10 \ = \ 0 $ and $ \ x_2^2 - 3 x_2 -10 \ = \ 0 \ \ . \ $ Subtracting the second equation from the first gives us $ \ x_1^2 - x_2^2 \ - \ 3·(x_1 - x_2) \ = \ 0 \ \Rightarrow \ x_1^2 - x_2^2 \ = \ 3·(x_1 - x_2) \ \ .^{*} \ $ Since the difference of the two (real) zeroes of a quadratic polynomial is $ \ | \ x_1 - x_2 \ | \ = \ \frac{\sqrt{\Delta}}{a} \ \ , \ \ (a \ = \ 1 \ $ here) with $ \ \Delta \ $ being the discriminant, it is straightforward to calculate the quantity sought. Because we are taking a difference, the order of the zeroes matters, so what we will obtain is $ \ | \ x_1^2 - x_2^2 \ | \ \ . \ $ (This problem is simple to check since the quadratic polynomial is easily factored.)

$ ^{*} $ Notice that this is a "disguised" way of writing $ \ x_1^2 - x_2^2 \ = \ (x_1 + x_2)·(x_1 - x_2) \ \ . $