2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I also do the same thing that you do $\endgroup$ – Abhishek Kumar May 4 at 5:41
  • 1
    $\begingroup$ $440$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.