How do I simplify this imaginary expression on the left?
x^2-(6+3i)x +k=0x2−(6+3i)x+k=0
one solution is 3
I solved for k
which is 9-9i9−9i (answer is correct)
But I am having issues factoring and simplifying when we plug in k in the original
x^2-(6+3i)x + (9-9i)=0x2−(6+3i)x+(9−9i)=0
one solution is 3
I solved for k
which is
But I am having issues factoring and simplifying when we plug in k in the original
1 Answer
Explanation:
Given:
x^2-(6+3i)x+k = 0" "x2−(6+3i)x+k=0 with root33
Putting
0 = color(blue)(3)^2-(6+3i)(color(blue)(3))+k0=32−(6+3i)(3)+k
color(white)(0) = 9-18-9i+k0=9−18−9i+k
color(white)(0) = -9-9i+k0=−9−9i+k
So:
k = 9 + 9ik=9+9i
Our original equation becomes:
x^2-(6+3i)x+(9+9i) = 0x2−(6+3i)x+(9+9i)=0
Note that:
6 + 3i = 3 + (3+3i)" "6+3i=3+(3+3i) which is the sum of the roots
9 + 9i = 3(3+3i)" "9+9i=3(3+3i) which is the product of the roots
Factoring, we have:
x^2-(6+3i)x+(9+9i) = (x-3)(x-(3+3i))x2−(6+3i)x+(9+9i)=(x−3)(x−(3+3i))
