Question

X^4-30x^2+125=0 solve the equation by making an approproate substitution

Answer 1

$x^4-30x^2+125=0 \\ \\t=x^2\\\\t^2-30t+125=0\\\\a=1, \ \ b=-30, \ \ t=125 \\ \\\Delta =b^2-4ac = (-30)^2 -4\cdot1\cdot 125 = 900-500=400$

$x^{2}=t \\ \\x^{2}=5 \ \ or \ \ x^2 = 25\\\\x^{2}-5=0 \ \ or \ \ x^2 - 25=0\\\\ (x-\sqrt{5}) (x+\sqrt{5})=0 \ \ \ or \ \ \(x-5)(x+5)=0 \\\\x=\sqrt{5}\ \ or \ \x= -\sqrt{5}\ \ or \ \ x=5 \ \ or \ \ x=-5 \\\\Answer : \ x=\left \{ -5,\ -\sqrt{5},\ \sqrt{5}, \ 5 \right \}$

Answer 2

$x^4-30x^2+125=0\\ \\ \boxed{y=x^2}\\ \\ y^2-30y+125=0\\ \\ \Delta=(-30)^2-4.1.125=900-500=400\\ \\ y=\frac{30 \pm20}{2}\\ \\ y_1=5\\ \\ y_2=25$

$x^2=5\Rightarrow x=\pm\sqrt5\\ \\ x_2=25 \Rightarrow x=\pm5\\ \\ S=\{-\sqrt5,\sqrt5,-5,5\}$