# 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?

Jan 24, 2017

#### Explanation:

Assume the quadratic function to be $y = a {x}^{2} + b x + c$.

Two real solutions

If it cuts $x$-axis at a point say $\left(h , 0\right)$, it means that when $x = h$, $y = 0$ and hence $x = h$ is a solution of quadratic equation $a {x}^{2} + b x + c = 0$. This happens when ${b}^{2} - 4 a c > 0$.

As $y = a {x}^{2} + b x + c = a \left({x}^{2} + \frac{b}{a} x\right) + c$

= $a \left({x}^{2} + 2 \times \frac{b}{2 a} x + {\left(\frac{b}{2} a\right)}^{2}\right) - {b}^{2} / \left(4 a\right) + c$

= $a {\left(x + \frac{b}{2 a}\right)}^{2} - \frac{{b}^{2} - 4 a c}{4 a}$ ................(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 ${b}^{2} - 4 a c = 0$. Observe that in (A), if it is so, we get $y = 0$ only when $x = - \frac{b}{2 a}$

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 ${b}^{2} - 4 a c < 0$. Observe that in (A), as ${b}^{2} - 4 a c < 0$, 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]}