# Solve the equation $\sqrt{x^2+x+1}+\sqrt{x^2+\frac{3x}{4}}=\sqrt{4x^2+3x}$

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?

• You forgot $-1$ before "Let's raise both sides to the power of 2 again but this time" Commented Oct 31, 2022 at 22:17
• @AnneBauval, yep, here goes my mistake. Thank you! Commented Oct 31, 2022 at 22:17
• 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. Commented Oct 31, 2022 at 22:58

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?

• Any reason for the downvote? Commented Oct 31, 2022 at 22:10
• Very clever! Thank you! Commented Oct 31, 2022 at 22:12
• @Kaloyan you are welcome! I am glad I could help. Commented Oct 31, 2022 at 22:14

Let me try. One can rewrite equation as below.

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

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

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

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

$$x = -4.$$

• 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! Commented Oct 31, 2022 at 22:15

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