The zeroes of the equation 4i,-4i,5,-5
Explanation:
Given:
[tex]-3x^4+27x^2+1200=0[/tex]
To Find:
The 0's of the equation=?
Solution:
We can write the equation by taking minus sign common from left hand side and the equation will become
[tex]3x^4-27 x^2-1200=0[/tex]
Now, Let[tex]x^2=t[/tex]
And put the value of [tex]x^2[/tex] in the above equation and then we will get
[tex]3t^2-27t-1200=0[/tex]
Now take 3 common from left hand side of the equation So equation would become
[tex]t^2-9t-400=0[/tex]
[tex](x+16)(x-25)=0[/tex]
hence the roots are t= -16 and 25
Now that we know the value(s) of t , we can calculate x since x is [tex]\sqrt{t}[/tex]
Since we are speaking 2nd root, each of the two imaginary solutions of has 2 roots
[tex]\sqrt{t}= x[/tex]
[tex] x=\pm\sqrt{-16}[/tex]
x= [tex]\pm4i[/tex]
[tex] x=\pm\sqrt{25}[/tex]
x= [tex]\pm5[/tex]
Hence the roots are 4i,-4i,5,-5
