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May 24th, 2018, 09:31 AM  #1 
Senior Member Joined: Jan 2012 Posts: 137 Thanks: 2  What's a pseudo quadratic equation?
Hi, What's pseudo quadratic equation? And it is solved? Thx. 
May 24th, 2018, 09:56 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,581 Thanks: 1038  
June 19th, 2018, 07:45 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
I had not seen the term "pseudoquadratic equation" before but it appears to be what I have seen called "an equation of quadratic type". That is any equation that can be made into a quadratic by a substitution. For example, the equation $\displaystyle x^4 6x^2+ 3x= 0$, while quartic, is "of quadratic type" because the substitution $\displaystyle y= x^2$ converts it to a quadratiic equation (in $\displaystyle y$). With that substitution, we have $\displaystyle y^2 6y+ 3= 0$. Completing the square, $\displaystyle y^2 6y+ 9 6= 0$, $\displaystyle (y 3)^2= 6$, $\displaystyle y= 3\pm\sqrt{6}$ so that $\displaystyle x^2j= 2\pm\sqrt{6}$ and the original equation has the four roots $\displaystyle x= \sqrt{2+ \sqrt{6}}$, $\displaystyle x= \sqrt{2+\sqrt{6}}$, $\displaystyle x= \sqrt{2 \sqrt{6}}$, and $\displaystyle x= \sqrt{2 \sqrt{6}}$. Similarly, $\displaystyle \sin^2(x)+ 3\sin(x) 4= 0$ is "of quadratic type" because the substitution $\displaystyle y= \sin(x)$ converts it to $\displaystyle y^2+ 3y 4= 0$, a quadratic equation. That can be factored as $\displaystyle (y+ 4)(y 1)$ so we have $\displaystyle y= \sin(x)= 4$ or $\displaystyle y= \sin(x)= 1$. Of course, $\displaystyle \sin(x)$ cannot equal 4, so the roots are given by $\displaystyle \sin(x)= 1$: $\displaystyle x= (2n+1)\frac{\pi}{2}$. Last edited by skipjack; June 19th, 2018 at 10:54 AM. 
June 19th, 2018, 10:51 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
The equation $\displaystyle x^4 6x^2+ 3x= 0$ isn't of quadratic type. I assume you intended to give $\displaystyle x^4  6x^2 + 3 = 0$. Its solutions are given by $\displaystyle x = \pm\sqrt{3 \pm \sqrt6}$, not $\displaystyle x = \pm\sqrt{2 \pm \sqrt6}$. The solution of $\sin(x) = 1$ is $\displaystyle x = (4n + 1)\frac\pi2$, not $\displaystyle x = (2n + 1)\frac\pi2$. 

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