5
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I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.

Can anyone provide a hint about what should be done?

  • 0
    Move one radical to the other side, and square both sides, then repeat. – 2017-01-13
  • 0
    I tried the same thing before posting the question, but could not completely solve it.... – 2017-01-13
  • 0
    You were on the right track. – 2017-01-13
  • 0
    @Lucian "Move one radical to the other side, and square both sides, then repeat." No thanks. "You were on the right track." Were they? Did you try to apply the approach you suggest, just to see how it works? – 2017-01-13
  • 0
    @Did: Works just fine. – 2017-01-13

4 Answers 4

7

Notice that

$$\left({\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}}\right) \left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) = 9x-3.$$ Thus,

$$9x-3 = \left(9x-3\right)\left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) \implies $$

$$9x-3 = 0 \implies x=1/3$$ or

$${\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}} = 1.$$

But the minimal value of $x^2-x+1$ is $3/4$, which implies that $$ 2{\sqrt {x^2-x+1}} \geq \sqrt{3} > 1.$$

Therefore $x =1/3$ is the only real solution.

  • 1
    Nice Job. +1 . Just one question, did you find the minimal value of $x^2 - x+1$ by differentiating ??? – 2017-01-13
  • 3
    Thanks :) No, I have just used the vertex coordinate $-\Delta/4a$. – 2017-01-13
  • 1
    @Alex Silva I really like your solution. So simple and elegant – 2017-01-13
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    @juniven, Thanks :) – 2017-01-13
4

Rewrite your equation so that there is only one square root on each side. We get $$\sqrt{4x^2+5x+1}=2\sqrt{x^2-x+1}+9x-3.$$ Squaring both sides we get $$ \begin{align} 4x^2+5x+1&=4(x^2-x+1)+4(9x-3)\sqrt{x^2-x+1}+(81x^2-54x+9)\\ &=85x^2-58x+13+4(9x-3)\sqrt{x^2-x+1}\end{align}.$$ We get $$81x^2-63x+12=-4(9x-3)\sqrt{x^2-x+1}.$$ Dividing by $3$, we get $$27x^2-21x+4=-4(3x-1)\sqrt{x^2-x+1}.$$ Apply factoring at the LHS, we get $$(9x-4)(3x-1)=-(3x-1)\cdot 4\sqrt{x^2-x+1}$$ Hence, $$(9x-4)(3x-1)+(3x-1)\cdot 4\sqrt{x^2-x+1}=0.$$ Thus, $$(3x-1)\cdot\Big[9x-4)+ 4\sqrt{x^2-x+1}\Big]=0.$$ Either $x=\frac{1}{3}$ or $$9x-4=-4\sqrt{x^2-x+1}.$$ Squaring again to both sides, we get $$81x^2-72x+16=16(x^2-x+1).$$ We get $$65x^2-56x=0.$$ Either $x=0$ or $x=\frac{56}{65}.$ Only $x=\frac{1}{3}$ satisfies the original equation.

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    You're right. However the downvote doesn't come from me :) – 2017-01-13
  • 1
    $x = \frac{1}{3}$ also works. You discarded it when you divided both sides by $3x -1$. – 2017-01-13
  • 0
    @Yiyuan Lee I will edit my answer thank you – 2017-01-13
2

Try to multiply both sides by $(\sqrt{4x^2+5x+1}+2\sqrt{x^2-x+1})$.

In the left side you will get $ 9x-3$ by difference of squares.

-1

Move $2\sqrt {x^2 - x + 1}$ to the rhs, then square both sides.