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

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.

• 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 – Eevee Trainer May 4 at 5:29
• I remove the radicals first then I do squaring so that I get the ans. – Abhishek Kumar May 4 at 5:31
• I don't think there is a "faster" way of solving this – Fareed AF May 4 at 5:42
• If we let $y = 2x^2-5x+2$, then we can get $x = 2y^2-5y+2$. Not sure if this helps. – Jerry Chang May 4 at 6:04
• Wonder how this was constructed. – 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:

$$y-x=2(x^2-y^2)-5(x-y)$$

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 $$x = \frac{3 - \sqrt{5}}{2}$$

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