Solve $2x^2-5x+2$= $\frac{5-\sqrt{9+8x}}{4}$

I simply do square both sides solve it and I get two value of x one is 2 and other is $\frac{3-√5}{2}$ but this approach it take more time so is there any approach for solving this equation.

  • $\begingroup$ I feel like the process would be slightly more expedient if you isolate the radical all to its own side before squaring but that's not much $\endgroup$ – Eevee Trainer May 4 at 5:29
  • $\begingroup$ I remove the radicals first then I do squaring so that I get the ans. $\endgroup$ – Abhishek Kumar May 4 at 5:31
  • $\begingroup$ I don't think there is a "faster" way of solving this $\endgroup$ – Fareed AF May 4 at 5:42
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    $\begingroup$ If we let $y = 2x^2-5x+2$, then we can get $x = 2y^2-5y+2$. Not sure if this helps. $\endgroup$ – Jerry Chang May 4 at 6:04
  • $\begingroup$ Wonder how this was constructed. $\endgroup$ – marty cohen May 4 at 6:27

Taking Jerry Chang's observation (which amounts to the fact that the expression on the right-hand side is what you get when you plug the coefficients of the quadratic into the quadratic formula, just choosing the minus sign for the square root) and setting $y=2x^2-5x+2$ so that $x=2y^2-5y+2$ we can subtract one of these from the other to obtain:


Which yields $y=x$ ; or

$1=5-2(x+y)$ ie $x+y=2$

Then the problem splits as $2x^2-5x+2=x$ or $2x^2-5x+2=2-x$

The solutions of these equations have to be plugged back into the original for checking to see which belongs to which choice of sign of the square root.


I don't know another way then going through the algebra, I got the same solutions. $$\begin{align} 2x^2-5x+2 &= \frac{5-\sqrt{9+8x}}{4} \\ 8x^2-20x+3 &= -\sqrt{9+8x} \\ (8x^2-20x+3)^2 &= (-\sqrt{9+8x})^2 \\ 64x^4-320x^3+448x^2-120x+9 &= 9 +8x\\ 64x^4-320x^3+440x^2-120x &= 0 \\ 64x(x-2)(x^2-3x+1) &=0 \\ \end{align}$$ We can see the solutions to that are $0$, $x=2$, $x = \frac{3 + \sqrt{5}}{2}$, $x = \frac{3 - \sqrt{5}}{2}$. Then plugging those into the original equation we get $x=2$ and $x = \frac{3 - \sqrt{5}}{2}$

  • $\begingroup$ I also do the same thing that you do $\endgroup$ – Abhishek Kumar May 4 at 5:41
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    $\begingroup$ $440$ and $120$ are not divisible by $64$ and $120$ is divisible by $5$, so your final factorisation is incorrect. $\endgroup$ – Mark Bennet May 4 at 6:46
  • $\begingroup$ @MarkBennet yeah i was like.....this is oddly clean, until i realized 440 couldn't possibly be divisible by 64 $\endgroup$ – Saketh Malyala May 4 at 6:47

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