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?

1 Answer
Jan 24, 2017

Answer:

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]}