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-√(5)) (x+√(5))=0 \ \ \ or \ \ \(x-5)(x+5)=0 \\\\x=√(5)\ \ or \ \x= -√(5)\ \ or \ \ x=5 \ \ or \ \ x=-5 \\\\Answer : \ x=\left \{ -5,\ -√(5),\ √(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=(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\}`