Answer:
[tex]x_{1}=5;x_{2}=-5;x_{3}=4;x_{4}=-4[/tex]
Step-by-step explanation:
The given expression is
[tex]x^{4}-41x^{2} =-400[/tex]
First, we have to move all terms to the left side of the equation
[tex]x^{4}-41x^{2}+400 =0[/tex]
Now, we can do a change of variable to transform this relation into a quadratic one. So
[tex]y^{2}=x^{4}; y=x^{2}[/tex]
Then, [tex]y^{2}-41y+400=0[/tex]
Now, we applied the quadratic formula
[tex]y_{1,2}=\frac{-b\±\sqrt{b^{2}-4ac} }{2a}[/tex]
Where
[tex]a=1;b=-41;c=400[/tex]
Replacing these values, we have
[tex]y_{1,2}=\frac{-(-41)\±\sqrt{(-41)^{2}-4(1)(400)} }{2(1)}\\y_{1,2}=\frac{41\±\sqrt{(1681-1600} }{2}=\frac{41\±\sqrt{81} }{2}\\ y_{1,2}=\frac{41\±9}{2}\\ y_{1}=\frac{41+9}{2}=25\\y_{2}=\frac{41-9}{2}=16[/tex]
However, we need to revert the variable change
[tex]x_{1} ^{2} =25\\\\x_{2} ^{2}=16\\x_{1}=\±5\\x_{2}=\±4[/tex]
This means that the expression only have real solution, which are
[tex]x_{1}=5;x_{2}=-5;x_{3}=4;x_{4}=-4[/tex]
