My problem is Solve the equation $$2 \left(\sqrt{x^3-7 x^2+17x-14}+\sqrt{x^4-7 x^3+23x^2-37 x+28}\right)=4x^2-17 x+25.$$ And my solve.

We have $$\sqrt{x^3-7 x^2+17x-14}= \sqrt{(x-2) \left(x^2-5 x+7\right)}\leqslant \dfrac{(x-2) + (x^2-5 x+7)}{2}= \dfrac{x^2 - 4x + 5}{2}.$$ Another way \begin{align*} \sqrt{x^4-7 x^3+23x^2-37 x+28} &= \sqrt{\left(x^2-4 x+7\right)\left(x^2-3 x+4\right)}\\ &\leqslant \dfrac{(x^2-4 x+7)+(x^2-3 x+4)}{2} \\ &= \dfrac{2x^2-7x+11}{2}. \end{align*} Therefore, $$\text{LHS} \leqslant (x^2 - 4x + 5) + (2x^2-7x+11) = 3x^2 -11x + 16.$$ From the given equation, we have $$4x^2-17 x+25 \leqslant 3x^2 -11x + 16 \Leftrightarrow x^2 -6x + 9 \leqslant 0 \leqslant (x-3)^2 \leqslant 0 \Leftrightarrow x = 3. $$ We see that, $x = 3$ satisfies the given equation.

Thus, the the given equation has only solution is $x = 3.$

Is there another way to solve this equation?


There is a very laborious way of doing it using successive squarings (which will create extra roots.

Starting with $$2(\sqrt A+\sqrt B)=C$$ a first squaring leads to $$4(A+B)-C^2=8\sqrt{AB}$$ and a second squaring leads to $$\Big(4(A+B)-C^2\Big)^2=64AB$$ Using $$A=x^3-7 x^2+17 x-14\qquad B=x^4-7 x^3+23 x^2-37 x+28\qquad C=4 x^2-17 x+25$$ this leads to $$144 x^8-2752 x^7+23640 x^6-119376 x^5+387945 x^4-831620 x^3+1149958 x^2-939876 x+348849=0$$ By inspection, this equation shows a double root $x=3$. Then, what is left is $$144 x^6-1888 x^5+11016 x^4-36288 x^3+71073 x^2-78590 x+38761=0$$ As Laplacian Fourier commented, this "last equation has no real roots by Descartes rule of signs".

  • 2
    $\begingroup$ I like the part where you say 'By inspection'. +1 for that. $\endgroup$
    – Shailesh
    Feb 25 '16 at 4:32
  • $\begingroup$ You can see thAt the last equation has no real roots by decartes rule of signs $\endgroup$
    – Teoc
    Feb 25 '16 at 4:36
  • $\begingroup$ @Shailesh. Is it humor ? I always look at polynomials using $x=0,\pm 1,\pm 2,\pm 3, \pm 4$ to see what happens. Cheers. $\endgroup$ Feb 25 '16 at 4:38
  • $\begingroup$ @LaplacianFourier. Thanks for pointing this key point. I shall edit quoting you. $\endgroup$ Feb 25 '16 at 4:39
  • $\begingroup$ I think, this is not a solution. $\endgroup$ Feb 25 '16 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.