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Say we have two quadratic equations as follows, $$x^2-2mx-1=0$$ $$x^2+(m+3)x-4=0$$ where $m$ is real, such that the two equations have one root that is the additive inverse of each other.

(For example, one may have root $x=3$ and the other would have $x=-3$ as one of its roots.)

Find m.


My only thought is using the quadratic formula and compare the two equation's roots, it'll get the job done, but takes a lot of work.

I suppose there's a method to modify one of the equations so that the new equation from modification has one same root as the equation we didn't modify.

That way, we can just subtract them and solve for a much simpler equation.

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1 Answer 1

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If $\alpha$ is a root of $f(x) = x^2 - 2mx - 1$, then $-\alpha$ is a root of $f(-x)$ or $x^2 + 2mx - 1$.

Now $x^2 + 2mx - 1$ and $x^2 + (m+3)x - 4$ must have a common root, or...

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  • $\begingroup$ Some algebra later, you get two values for $m$. Check that both values of $m$ are valid. $\endgroup$
    – Toby Mak
    Jan 22 at 8:07
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    $\begingroup$ $m = \frac{24}{7}$ and $m = 0$, both should be valid $\endgroup$
    – Cyh1368
    Jan 22 at 8:26
  • $\begingroup$ Yep, that's correct! $\endgroup$
    – Toby Mak
    Jan 22 at 9:17

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