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 May 24th, 2018, 09:31 AM #1 Senior Member     Joined: Jan 2012 Posts: 140 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: 14,597 Thanks: 1038 Thanks from happy21, Country Boy and JeffM1
 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 "pseudo-quadratic 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: 20,978 Thanks: 2229 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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