# Can anyone find the roots of this quadratic equation?

1. Dec 16, 2017

### loveinla

Here a, b, c > 0, and a > bc.

Can anyone find the solution of k as a function of (a, b, c)? Thanks.

2. Dec 16, 2017

### Staff: Mentor

This is not a quadratic equation. Where does it come from? It is a quartic (with ^4 terms) equation in $\sqrt{k}$ so technically there is a closed solution but that is really messy and beyond the scope of homework problems.
Can you post the full original problem and how you got to this point?

3. Dec 17, 2017

### epenguin

Yes mfb probably has in mind, no one would invent that expression for an academic exercise, it presumably comes out of some problem of geometry or something else. And even then when you get expressions like that it is often a sign that you made a mistake in deriving them.

That said, if it is really that or something similar you can least note that b and c never appear separately, but only as bc, so you might as well replace that by a single symbol, d, say. Then notice you have a square root of something complicated, but it is the same thing in three terms. So the procedure would be to combine them as just the square root multiplied by something,Then ‘take that to the other side’ (hope you know what that means) then square both sides, then see if any cancellations or other simplifications are possible. But by the looks of it I would expect that to give a mess as mfb says.

4. Dec 17, 2017

### Ray Vickson

The solution is so horrible as to be essentially unusable. Maple manages to find four roots, by finding and solving the equivalent fourth-degree polynomial. One of the roots is
r1 =
-15/58*b*c+15/29*a+1/2*(1/4*(30/29*b*c-60/29*a)^2-44/87*b^2*c^2+40/29*a*b*c-26/29*a^2+1/87*(1126*b^6*c^6-3132*a*b^5*c^5+1170*a^2*b^4*c^4+3582*a^3*b^3*c^3-2772*a^4*b^2*c^2-1026*a^5*b*c+1053*a^6+6*(-3807*b^12*c^12+67500*a*b^11*c^11-460278*a^2*b^10*c^10+1562622*a^3*b^9*c^9-2919687*a^4*b^8*c^8+3050688*a^5*b^7*c^7-1736700*a^6*b^6*c^6+783348*a^7*b^5*c^5-884757*a^8*b^4*c^4+841092*a^9*b^3*c^3-314046*a^10*b^2*c^2-7650*a^11*b*c+21675*a^12)^(1/2))^(1/3)-87*(-112/7569*b^4*c^4+28/841*a*b^3*c^3-68/2523*a^2*b^2*c^2+44/2523*a^3*b*c-23/2523*a^4)/(1126*b^6*c^6-3132*a*b^5*c^5+1170*a^2*b^4*c^4+3582*a^3*b^3*c^3-2772*a^4*b^2*c^2-1026*a^5*b*c+1053*a^6+6*(-3807*b^12*c^12+67500*a*b^11*c^11-460278*a^2*b^10*c^10+1562622*a^3*b^9*c^9-2919687*a^4*b^8*c^8+3050688*a^5*b^7*c^7-1736700*a^6*b^6*c^6+783348*a^7*b^5*c^5-884757*a^8*b^4*c^4+841092*a^9*b^3*c^3-314046*a^10*b^2*c^2-7650*a^11*b*c+21675*a^12)^(1/2))^(1/3))^(1/2)-1/2*(1/2*(30/29*b*c-60/29*a)^2-88/87*b^2*c^2+80/29*a*b*c-52/29*a^2-1/87*(1126*b^6*c^6-3132*a*b^5*c^5+1170*a^2*b^4*c^4+3582*a^3*b^3*c^3-2772*a^4*b^2*c^2-1026*a^5*b*c+1053*a^6+6*(-3807*b^12*c^12+67500*a*b^11*c^11-460278*a^2*b^10*c^10+1562622*a^3*b^9*c^9-2919687*a^4*b^8*c^8+3050688*a^5*b^7*c^7-1736700*a^6*b^6*c^6+783348*a^7*b^5*c^5-884757*a^8*b^4*c^4+841092*a^9*b^3*c^3-314046*a^10*b^2*c^2-7650*a^11*b*c+21675*a^12)^(1/2))^(1/3)+87*(-112/7569*b^4*c^4+28/841*a*b^3*c^3-68/2523*a^2*b^2*c^2+44/2523*a^3*b*c-23/2523*a^4)/(1126*b^6*c^6-3132*a*b^5*c^5+1170*a^2*b^4*c^4+3582*a^3*b^3*c^3-2772*a^4*b^2*c^2-1026*a^5*b*c+1053*a^6+6*(-3807*b^12*c^12+67500*a*b^11*c^11-460278*a^2*b^10*c^10+1562622*a^3*b^9*c^9-2919687*a^4*b^8*c^8+3050688*a^5*b^7*c^7-1736700*a^6*b^6*c^6+783348*a^7*b^5*c^5-884757*a^8*b^4*c^4+841092*a^9*b^3*c^3-314046*a^10*b^2*c^2-7650*a^11*b*c+21675*a^12)^(1/2))^(1/3)+((22/29*b^2*c^2-60/29*a*b*c+39/29*a^2)*(30/29*b*c-60/29*a)-16/29*b^3*c^3+52/29*a*b^2*c^2-56/29*a^2*b*c+20/29*a^3-1/4*(30/29*b*c-60/29*a)^3)/(1/4*(30/29*b*c-60/29*a)^2-44/87*b^2*c^2+40/29*a*b*c-26/29*a^2+1/87*(1126*b^6*c^6-3132*a*b^5*c^5+1170*a^2*b^4*c^4+3582*a^3*b^3*c^3-2772*a^4*b^2*c^2-1026*a^5*b*c+1053*a^6+6*(-3807*b^12*c^12+67500*a*b^11*c^11-460278*a^2*b^10*c^10+1562622*a^3*b^9*c^9-2919687*a^4*b^8*c^8+3050688*a^5*b^7*c^7-1736700*a^6*b^6*c^6+783348*a^7*b^5*c^5-884757*a^8*b^4*c^4+841092*a^9*b^3*c^3-314046*a^10*b^2*c^2-7650*a^11*b*c+21675*a^12)^(1/2))^(1/3)-87*(-112/7569*b^4*c^4+28/841*a*b^3*c^3-68/2523*a^2*b^2*c^2+44/2523*a^3*b*c-23/2523*a^4)/(1126*b^6*c^6-3132*a*b^5*c^5+1170*a^2*b^4*c^4+3582*a^3*b^3*c^3-2772*a^4*b^2*c^2-1026*a^5*b*c+1053*a^6+6*(-3807*b^12*c^12+67500*a*b^11*c^11-460278*a^2*b^10*c^10+1562622*a^3*b^9*c^9-2919687*a^4*b^8*c^8+3050688*a^5*b^7*c^7-1736700*a^6*b^6*c^6+783348*a^7*b^5*c^5-884757*a^8*b^4*c^4+841092*a^9*b^3*c^3-314046*a^10*b^2*c^2-7650*a^11*b*c+21675*a^12)^(1/2))^(1/3))^(1/2))^(1/2)
and there are three more like that.

And, of course, one problem is that the process of converting the problem go a polynomial-solving task can introduce extraneous solutions---solutions of the polynomial that are not solutions of the original equation. Telling which of the four symbolic solutions fails to solve the original equation is just about impossible with expressions as complicated as the above. However, looking at some numerical examples, we can just plot a graph of your $f(k)$ and see how many real roots it has. For example, when $a=3, b=1, c=2$ (which satisfies your restriction $a > bc$) we have the equation $F(k) = 0$, where
$$F(k) = 2-6k+19k^2 + (2-7k) \sqrt{k(5k+8)}$$
The graph of $F(k)$ crosses zero in just two points, $k = 0.542990634$ and $k = 3.371514666$. The other two (complex) roots of the 4th degree polynomial are decidedly not roots of the original, unconverted equation---not even complex roots of $F(k)$.

Last edited: Dec 17, 2017
5. Dec 17, 2017

### Staff: Mentor

6. Dec 17, 2017

### FactChecker

I'm not sure that I would call that useless. A computer doesn't get tired.

7. Dec 17, 2017

### Staff: Mentor

Sure, but you can ask a computer to get numerical solutions without that expression as well.

I said "really messy", but that expression is even worse than I expected.

8. Dec 17, 2017

### FactChecker

I agree. I see some large powers like a^12. I was half joking about the computer not getting tired, but not really sure if that formula would be useful or not..

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