There could be five in each class.
X-y=3=>x=3+y
X+2y=-6
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3+y+2y=-6
3+3y=-6
3y=-6-3
3y=-9
y=-3
X+2y=-6
X+2(-3)=-6
X-6=-6
X=0
Answer:
The roots are;
x = (2 + i)/5 or (2-i)/5
where the term i is the complex number representing the square root of -1
Step-by-step explanation:
Here, we want to use the completing the square method to solve the quadratic equation;
f(x) = -5x^2 + 4x -1
Set the function to zero
0 = -5x^2 + 4x - 1
So;
-5x^2 + 4x = 1
divide through by the coefficient of x which is -5
x^2 - 4/5x = -1/5
We now take half of the coefficient of x and square it
= -2/5^2 = 4/25
add it to both sides
x^2 - 4x/5 + 4/25= -1/5 + 4/25
(x- 2/5)^2 = -1/5 + 4/25
(x - 2/5)^2 = -1/25
Take the square root of both sides
x - 2/5 = √( -1/25
x - 2/5 = +i/5 or -i/5
x = 2/5 + i/5 or 2/5 - i/5
The answer is D to the next step
