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