Question

How do you determine whether there are two, one or no real solutions given the graph of a quadratics function intersects the x axis twice?

Answer 1

Please see below.

Explanation:

Assume the quadratic function to be `y=ax^2+bx+c`.

Two real solutions

If it cuts `x`-axis at a point say `(h,0)`, it means that when `x=h`, `y=0` and hence `x=h` is a solution of quadratic equation `ax^2+bx+c=0`. This happens when `b^2-4ac>0`.

As `y=ax^2+bx+c=a(x^2+b/ax)+c`

= `a(x^2+2xxb/(2a)x+(b/2a)^2)-b^2/(4a)+c`

= `a(x+b/(2a))^2-(b^2-4ac)/(4a)` ................(A)

there could be two values of `x`, for which `y=0`.

Example: Shows two graphs each for different parameters. While one opens upwards other opens downward.
graph{(y-3x^2+7x-3)(y+x^2+2x-1)=0 [-3.854, 6.146, -2.66, 2.34]}

One real solution

In such a case the quadratic function touches `x`-axis and does not cuts `x`-axis and hence there is just one solution. This happens when `b^2-4ac=0`. Observe that in (A), if it is so, we get `y=0` only when `x=-b/(2a)`

Examples: graph{(y-x^2+2x-1)(y+x^2+4x+4)=0 [-3.854, 6.146, -2.66, 2.34]}

No real solution

In such a case the quadratic function does not touch / cut `x`-axis at all and hence there is no solution. This happens when `b^2-4ac<0`. Observe that in (A), as `b^2-4ac<0`, we have `y>0` or `y<0` depensing on `a`. If `a>0`, `y>0` and if `a<0`, `y<0`.

Examples: graph{(y-x^2-3x-3)(y+x^2-3x+3)=0 [-3.854, 6.146, -2.66, 2.34]}