Respuesta :
Answer:
There are four roots of [tex]p[/tex],
[tex]p= 3+\sqrt{2}i\\p = 3-\sqrt{2}i\\p= 3+\sqrt{10}\\p = 3-\sqrt{10}[/tex]
Step-by-step explanation:
Given:
[tex](p^2-6p)^2+10(p^2-6p)=11\\[/tex]
Let [tex]p^2-6p =x[/tex]
Now solving the equation we get:
[tex]x^2+10x=11\\x^2+10x-11=0\\x^2-x+11x-11=0\\x(x-1)+11(x-1)=0\\(x+11)(x-1)=0[/tex]
Now, solving for two values of [tex]x[/tex], we get;
[tex]x+11=0\\x=-11\\also; x-1=0\\x=1[/tex]
Now re-substituting value of [tex]x[/tex], we get;
[tex]p^2-6p=-11\\p^2-6p+11=0\\also p^2-6p=1\\p^2-6p-1=0\\[/tex]
Now we have two quadratic equation we solve for each,
1) We will solve for [tex]p^2-6p+11=0[/tex]
Using quadratic formula to solve it. This gives,
[tex]x= \frac{-b\±\sqrt{b^2-4ac}} {2a}[/tex]
Here [tex]b= -6, a=1, c =11[/tex]
hence,[tex]p= \frac{-(-6)\±\sqrt{(-6)^2-4\times 1\times 11}} {2\times 1}=\frac{6\±\sqrt{(36-44)}} {2}=\frac{6\±\sqrt{-8}} {2}= \frac{6\±2\sqrt{2}i} {2}[/tex]
[tex]p = \frac{6\±2\sqrt{2}i} {2}= 3\±\sqrt{2}i\\hence, p= 3+\sqrt{2}i \ or \ p = 3-\sqrt{2}i[/tex]
2) We will solve for [tex]p^2-6p-1=0[/tex]
Using quadratic formula to solve it. This gives,
[tex]x= \frac{-b\±\sqrt{b^2-4ac}} {2a}[/tex]
here [tex]b= -6, a=1, c =-1[/tex]
[tex]\therefore p= \frac{-(-6)\±\sqrt{(-6)^2-4\times 1\times -1}} {2\times 1}=\frac{6\±\sqrt{(36+4)}} {2}=\frac{6\±\sqrt{40}} {2}= \frac{6\±2\sqrt{10}} {2}[/tex]
[tex]p = \frac{6\±\sqrt{10}} {2}= 3\±\sqrt{10}\\hence, p= 3+\sqrt{10} \ or \ p = 3-\sqrt{10}[/tex]
Hence, from 1 and 2, we find there are 2 real roots and 2 imaginary roots.
[tex]p= 3+\sqrt{2}i\\p = 3-\sqrt{2}i\\p= 3+\sqrt{10}\\p = 3-\sqrt{10}[/tex]
