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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?

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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".

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    $\begingroup$ 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. $\endgroup$ – Claude Leibovici Feb 25 '16 at 4:38
  • $\begingroup$ @LaplacianFourier. Thanks for pointing this key point. I shall edit quoting you. $\endgroup$ – Claude Leibovici Feb 25 '16 at 4:39
  • $\begingroup$ I think, this is not a solution. $\endgroup$ – minthao_2011 Feb 25 '16 at 5:20

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