Question

show that the equation x⁴+15x²+7x-11=0 has one positive, one negative and two imaginary roots​

Answer 1

Answer:

x≓−1.070190132

Step-by-step explanation:

Answer 2

Answer:

x^4-4x^3+8x^2-8x+4 =0 can be written as

x^4-4x^3+(4x^2+4x^2)-8x+4=0 {8x^2=4x^2+4x^2}

Now group the part of the equation in this manner.

x^2(x^2-4x+4)+4(x^2-2x+1)=0

x^2(x-2)^2+4(x-1)^2=0

we can see that (x(x-2))^2 and (2(x-1))^2 would always be positive.

So to get the sum equal to zero we need both the parts to be equal to zero simultaneously. But this cannot be possible in this case so real roots do not exist in this case, we have to find imaginary roots.

x^2(x-2)^2=-4(x-1)^2

(x^2(x-2)^2)/(x-1)^2=-4

((x^2-2x)/x-1)^2=-4

((x^2-2x)/x-1)=+-2i

solve for this equations