Question

Determine the roots of p(x)=3(x+1)^4-5(x+1)^2+1"Show your work"

Answer 1

$p(x)=3(x+1)^4-5(x+1)^2+1=3\left[(x+1)^2\right]^2-5(x+1)^2+1\\\\\text{subtitute:}\ (x+1)^2=t\geq0$

$p(t)=3t^2-5t+1\\\\p(t)=0\iff3t^2-5t+1=0\\\\a=3;\ b=-5;\ c=1\\\\\Delta=b^2-4ac\to\Delta=(-5)^2-4\cdot3\cdot1=25-12=13 > 0\\\\\sqrt\Delta=\sqrt{13}\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-(-5)-\sqrt{13}}{2\cdot3}=\dfrac{5-\sqrt{13}}{6} > 0\\\\x_2=\dfrac{-(-5)+\sqrt{13}}{2\cdot3}=\dfrac{5+\sqrt{13}}{6} > 0$

$(x+1)^2=\dfrac{5-\sqrt{13}}{6}\ \vee\ (x+1)^2=\dfrac{5+\sqrt{13}}{6}\\\\x+1=\pm\sqrt{\dfrac{5-\sqrt{13}}{6}}\ \vee\ x+1=\pm\sqrt{\dfrac{5+\sqrt{13}}{6}}$

$Answer:\\\\\boxed{x\in\left\{-\sqrt{\dfrac{5-\sqrt{13}}{6}}-1;\ -\sqrt{\dfrac{5+\sqrt{13}}{6}}-1;\ \sqrt{\dfrac{5-\sqrt{13}}{6}}-1;\ \sqrt{\dfrac{5+\sqrt{13}}{6}}-1\right\}}$