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Solve the following equation:

$\sqrt{4x^2-15x+20}=4x-10+7\sqrt{x-1}$

Attempt:

$\sqrt{4x^2-15x+20}=4x-10+7\sqrt{x-1}$

$\Leftrightarrow {\begin{cases}\left(\sqrt{4x^2-15x+20}-7\sqrt{x-1}\right)\left(4x-10\right)\ge 0\\16x^2-80x+100=4x^2-15x+20+49x-49-14\sqrt{\left(4x^2-15x+20\right)\left(x-1\right)}\end{cases}}$

$\Leftrightarrow {\begin{cases}\left(\sqrt{4x^2-15x+20}-7\sqrt{x-1}\right)\left(4x-10\right)\ge 0\\14\sqrt{4x^3-19x^2+35x-20}=-12x^2+114x-129\end{cases}}$

$\Leftrightarrow {\begin{cases}\left(\sqrt{4x^2-15x+20}-7\sqrt{x-1}\right)\left(4x-10\right)\ge 0\\-12x^2+114x-129\ge 0\\196\left(4x^3-19x^2+35x-20\right)=\left(-12x^2+114x-129\right)^2\end{cases}}$

$\Leftrightarrow {\begin{cases}\left(\sqrt{4x^2-15x+20}-7\sqrt{x-1}\right)\left(4x-10\right)\ge 0\\-12x^2+114x-129\ge 0\\144x^4-3520x^3+19816x^2-36272x+20561=0\end{cases}}$

$\Leftrightarrow {\begin{cases}\left(\sqrt{4x^2-15x+20}-7\sqrt{x-1}\right)\left(4x-10\right)\ge 0\\-12x^2+114x-129\ge 0\\\left(4x^2-24x+29\right)\left(36x^2-664x+709\right)=0\end{cases}}$

$\Leftrightarrow x=\frac{6-\sqrt{7}}{2}$, which is correct.

Is there any way other than brute force to solve this equation? I do know some special methods to solve irrational equations, but I can't (or don't know how to) apply it here. The attempt above is my "traditional method", change it to another equation that does not contain square roots to solve it.

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  • $\begingroup$ Both members represent hyperbolas, which have potentially four intersections. Hence the problem can be turned to a quartic equation. $\endgroup$
    – user65203
    Mar 1, 2018 at 17:40

1 Answer 1

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Let $f(x)=4x-10+7\sqrt{x-1}-\sqrt{4x^2-15x+20}.$

Thus, the domain gives $x\geq1$ and since $$4x^2-15x+20=(x-1)^2+3x^2-13x+19>(x-1)^2,$$ we obtain: $$f'(x)=4-\frac{8x-15}{2\sqrt{4x^2-15x+20}}+\frac{7}{2\sqrt{x-1}}>\frac{8\sqrt{4x^2-15x+20}-(8x-15)}{2\sqrt{4x^2-15x+20}}>$$ $$>\frac{8(x-1)-(8x-15)}{2\sqrt{4x^2-15x+20}}=\frac{7}{2\sqrt{4x^2-15x+20}}>0,$$ which says that $f$ increases and $f$ has one root maximum.

But, $$f\left(3-\frac{\sqrt7}{2}\right)=12-2\sqrt7-10+7\sqrt{2-\frac{\sqrt{7}}{2}}-\sqrt{(6-\sqrt7)^2-15\left(3-\frac{\sqrt7}{2}\right)+20}=$$ $$=2-2\sqrt7+\frac{7}{2}\sqrt{8-2\sqrt7}-\sqrt{18-\frac{9}{2}\sqrt7}=2-2\sqrt7+\frac{7}{2}\sqrt{8-2\sqrt7}-\frac{3}{2}\sqrt{8-2\sqrt7}=$$ $$=2-2\sqrt7+2\sqrt{(\sqrt7-1)^2}=0,$$ which says that $3-\frac{\sqrt7}{2}$ is an unique root.

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