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Solve the equation $$\sqrt{x^2+x+1}+\sqrt{x^2+\dfrac{3x}{4}}=\sqrt{4x^2+3x}$$ The domain is $$x^2+\dfrac{3x}{4}\ge0,4x^2+3x\ge0$$ as $x^2+x+1>0$ for every $x$. Let's raise both sides to the power of 2: $$x^2+x+1+x^2+\dfrac{3x}{4}+2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=4x^2+3x\\2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}$$ Let's raise both sides to the power of 2 again but this time the roots should also satisfy $A:2x^2+\dfrac54x\ge0$:$$4(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)=(2x^2+\dfrac54x)^2$$ I came at $$x(2x^2+\dfrac{87}{16}x+3)=0$$ I obviously made a mistake as the answer is $x=-4$, but is there an easier approach?

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    $\begingroup$ You forgot $-1$ before "Let's raise both sides to the power of 2 again but this time" $\endgroup$ Commented Oct 31, 2022 at 22:17
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    $\begingroup$ @AnneBauval, yep, here goes my mistake. Thank you! $\endgroup$ Commented Oct 31, 2022 at 22:17
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    $\begingroup$ Note: Some care is generally required when squaring equations. Squaring $a=b$ to obtain $a^2=b^2$ has the effect of incorporating the solutions of $a=-b$ into the squared equation. So it is always wise, once a solution is found, to check that it solves the original equation. Not so much a problem with this one, but something to bear in mind. $\endgroup$ Commented Oct 31, 2022 at 22:58

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HINT

I would start with multiplying both sides by the number $2$: \begin{align*} \sqrt{x^{2} + x + 1} + \sqrt{x^{2} + \frac{3x}{4}} = \sqrt{4x^{2} + 3x} & \Longleftrightarrow 2\sqrt{x^{2} + x + 1} + \sqrt{4x^{2} + 3x} = 2\sqrt{4x^{2} + 3x}\\\\ & \Longleftrightarrow 2\sqrt{x^{2} + x + 1} = \sqrt{4x^{2} + 3x}\\\\ \end{align*}

Can you take it from here?

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  • $\begingroup$ Any reason for the downvote? $\endgroup$ Commented Oct 31, 2022 at 22:10
  • $\begingroup$ Very clever! Thank you! $\endgroup$ Commented Oct 31, 2022 at 22:12
  • $\begingroup$ @Kaloyan you are welcome! I am glad I could help. $\endgroup$ Commented Oct 31, 2022 at 22:14
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Let me try. One can rewrite equation as below.

$$\sqrt{x^2+x+1} + \frac{1}{2}\sqrt{4x^2+3x} = \sqrt{4x^2+3x},$$

$$\sqrt{x^2+x+1} = \frac{1}{2}\sqrt{4x^2+3x},$$

$$x^2+x+1 = \frac{1}{4}(4x^2+3x)$$

$$4x^2+4x+4 = 4x^2+3x,$$

$$x = -4.$$

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  • $\begingroup$ Thank you! Your approach is actually the same as in the other post. In the second-to-last line in your solution the LHS should be $+4$ as we have multiplied by $4$. Thank you! $\endgroup$ Commented Oct 31, 2022 at 22:15
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following your original approach you should have gotten in line 4:

$$2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}-1$$

and, squaring both sides and collecting like terms, yields

$32x^3+151x^2+88x-16=0$

which can be factored as

$(x+4)(32x^2+23x-4)=0$

from which you get $x=-4$ and two extraneous irrational roots

However,as already noted in other answers, this isn't the simplest approach

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