Thanks so much for your help. Didn't understand it at first so took a break from it and had a look at it again and all makes sense now.
When you substitute k= -√3 into y = x^2 + 2kx +3, you get y= x^2 - (2√3)x + 3.
The equation y= x^2 - (2√3)x + 3 has one real repeated root and that is x = √3. This is because when you solve for x (so sub y = 0 into y= x^2 - (2√3)x + 3), you get x= √3.
So the graph of y= x^2 - (2√3)x + 3 touches the x axis at x = √3.
When you substitute k= √3 into y = x^2 + 2kx +3, you get y= x^2 + (2√3)x + 3.
The equation y= x^2 + (2√3)x + 3 has one real repeated root and that is x = -√3. This is because when you solve for x (so sub y = 0 into y= x^2 + (2√3)x + 3), you get x= √3.
So the graph of y= x^2 + (2√3)x + 3 touches the x axis at x = -√3.
So you can conclude that when k = -√3 or k = √3, y = x^2 + 2kx +3 will have one repeated root.When you substitute a value of k that is between -√3 and √3 (so when -√3 <k< √3) , you should get an error when solving for x.
For example, when you substitute k = 0 into y = x^2 + 2kx +3, you get y= x^2 + 3.
The equation y= x^2 + 3 has no real roots as it does not cross the x axis (y = 0).This is because when you solve for x (so sub y = 0 into y= x^2 + 3), you get x^2 = -3. You cannot square root a negative number, so you don't get a value for x when y = 0.
So the graph of = x^2 + 2kx +3 is
above the x axis (
since x^2 is positive - anything you square will give a positive value) when k = 0, it does not intersect the x axis.
You could try another value of k that is between -√3 and √3. When we sub k = 1 into y = x^2 + 2kx +3 and use the quadratic formula to solve for x, we get x = (-2 + √-8)/2 or x = (-2 - √-8)/2. This generates an error because we cannot square root a negative number, in this case -8.
So you can conclude that when -√3 <k< √3, y = x^2 + 2kx +3 will have no real roots.