I love being Inspired (Part-4)

Algebra Level 4

x 3 + 6 x 2 + 4 x + 1 = 0 \large{x^{3}+6x^{2}+4x+1=0}

Find the only real solution of x x in the equation above, correct to 3 places of decimals.


The answer is -5.278.

Akshay Yadav by Akshay Yadav , 20, India

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1 solution

Gian Sanjaya
Sep 27, 2015

I could say that this is one hard problem if you don't tackle it with the help of a calculation engine.

We have x 3 + 6 x 2 + 4 x + 1 = ( x + 2 ) 3 − 8 ( x + 2 ) + 9 x^3+6x^2+4x+1 = (x+2)^3-8(x+2)+9 . Define y = x + 2 y = x+2 , so the equation becomes y 3 − 8 y + 9 = 0 y^3-8y+9=0 . The surely real root z of a cubic equation z 3 + a z + b = 0 z^3+az+b=0 is (I left it to be proved by someone, or correcting it if it's wrong):

z = ( a 3 ) 3 + ( b 2 ) 2 − b 2 3 − ( a 3 ) 3 + ( b 2 ) 2 + b 2 3 z = \sqrt[3]{\sqrt{(\frac{a}{3})^3+(\frac{b}{2})^2}-\frac{b}{2}} - \sqrt[3]{\sqrt{(\frac{a}{3})^3+(\frac{b}{2})^2}+\frac{b}{2}}

We have only one real root if and only if ( a 3 ) 3 + ( b 2 ) 2 > 0 (\frac{a}{3})^3+(\frac{b}{2})^2>0 (same, I left it to be proved by someone else, this is what I've found and my proof is just too long). We have it on our equation. Applying it on our equation gives this:

y = 417 − 81 18 3 − 417 + 81 18 3 ≈ − 3.278 y = \sqrt[3]{\frac{\sqrt{417}-81}{18}}-\sqrt[3]{\frac{\sqrt{417}+81}{18}} \approx -3.278

Hence, x = y − 2 ≈ − 5.278 x = y-2 \approx \boxed{-5.278} .

Note: I would suggest, if ( a 3 ) 3 + ( b 2 ) 2 < 0 (\frac{a}{3})^3+(\frac{b}{2})^2<0 , to take the cube root as the one with the imaginary part nearest possible to 0, but if b=0, sure we have z=0.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

The proof is just Cardano's method for the cubic equation.

Start with y 3 + a y + b = 0 y^3+ay+b=0 , then use the identity ( u + v ) 3 − 3 u v ( u + v ) − ( u 3 + v 3 ) = 0 (u+v)^3-3uv(u+v)-(u^3+v^3)=0 . Compare it with the equation to get:

y = u + v 3 u v = − a u 3 + v 3 = − b y=u+v \\ 3uv=-a \\ u^3+v^3=-b

We can divide the second equation by 3 and cube it to get: ( u v ) 3 = − ( a 3 ) 3 (uv)^3=-(\frac{a}{3})^3 . Then, by Vieta's formulas, u 3 u^3 and v 3 v^3 are roots of:

z 2 + b z − ( a 3 ) 3 = 0 z^2+bz-(\frac{a}{3})^3=0

By the quadratic formula we get:

z = − b 2 ± ( b 2 ) 2 + ( a 3 ) 3 z=-\frac{b}{2}\pm\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3}

So, we have the values for u u and v v (note that they are interchangeable):

u = − b 2 + ( b 2 ) 2 + ( a 3 ) 3 3 v = − b 2 − ( b 2 ) 2 + ( a 3 ) 3 3 u=\sqrt[3]{-\frac{b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3}} \\ v=\sqrt[3]{-\frac{b}{2}-\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3}}

Finally, y = − b 2 + ( b 2 ) 2 + ( a 3 ) 3 3 + − b 2 − ( b 2 ) 2 + ( a 3 ) 3 3 y=\sqrt[3]{-\frac{b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3}}+\sqrt[3]{-\frac{b}{2}-\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3}}

The expression inside the square root is − Δ 108 -\frac{\Delta}{108} , where Δ \Delta is the discriminant for this equation. If Δ > 0 \Delta>0 we have three real roots, if Δ = 0 \Delta=0 we have at least two multiple roots, and if Δ < 0 \Delta<0 we have one real root and two complex conjugate roots.

Alan Enrique Ontiveros Salazar - 5 years, 7 months ago

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