You can put this solution on YOUR website!
I'm not sure I understand the question.
If you want to solve these equations simultaneously, then you would have:
y = 3
y = x^2 - 6x + c
you would substitute 3 for y in the second equation to get:
3 = x^2 - 6x + c
If you subtract 3 from both sides of this equation, you would get:
x^2 - 6x + c - 3 = 0
If you consider a solution to this equation as finding the roots of this equation, then you can get no roots, one root, or two roots.
There are several ways to find the roots of this equation.
One way is to use the quadratic formula of
If the number under the square root sign is 0, you will have one root.
If the number under the square root sign is positive, you will have two roots.
If the number under the square root sign is negative, you will have no roots.
The standard form of a quadratic equation is:
y = ax^2 + bx + c = 0
Your equation is:
y = x^2 - 6x + c - 3 = 0
The a term is equal to 1
The b term is equal to -6
The c term is equal to c-3
To get one root, you want b^2 - 4ac to equal to 0
(-6)^2 - 4*1*(c-3)
Expanding this out, you have:
36 - 4c + 12 = 48-4c
Set this equal to 0 and you have:
48 - 4c = 0 which becomes 4c = 48 which becomes c = 12
If c = 12, you will have one root.
Your original equation becomes:
3 = x^2 - 6x + 12
Subtract 3 from both sides to get x^2 - 6x + 9 = 0
Factor this to get:
(x-3)^2 = 0
Take the square root of both sides of this equation to get x-3 = 0.
Add 3 to both sides of this equation to get x = 3
Your one root is x = 3
A graph of x^2 - 6x + 9 is shown below:
, then if you make c bigger, it will become negative and if you make c smaller, it will become positive.
to get two roots, make c smaller.
Something like 6 would be fine.
A graph of x^2 - 6x + 6 is shown below:
As you can see the equation shown above has two roots.
to get no roots, make c larger.
Something like 12 would be fine.
A graph of x^2 - 6x + 12 is shown below:
As you can see the equation shown above has no roots because it does not cross the x-axis where the roots would be.
You actually get solutions when you solve for the roots, but the solut9ions are not real because they contain imaginary numbers which are products of the square root of a negative number. These are not considered roots.
To find the roots of the second and third example, just plug the numbers into the quadratic formula and you'll get them.
Your answer is:
You'll get one solution when c = 9.
You'll get two solutions when c < 9.
You'll get no solutions when c > 9.