Question

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400

Answer 1

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

$x_{1}=5;x_{2}=-5;x_{3}=4;x_{4}=-4$

Answer 2

We are given the polynomial equation x^4 - 41x^2 + + 1400 and is asked to determine the roots of the equation whether real or complex. The standard form should be the equal to the equation where -400 is up in the left side. The answers are 4 integers: 2 are negative