# Question #3c005

Aug 11, 2016

There are 2 Real roots

#### Explanation:

Consider the standard form equation: $y = a {x}^{2} + b x + c$

Where: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Now consider the discriminate: ${b}^{2} - 4 a c = 10$

The fact that this is positive means that the solution for $x$ belongs to the set of Real Numbers: $x \in \mathbb{R}$

Thus we have $\frac{- b}{2 a} \pm \frac{\sqrt{10}}{2 a}$

Also $\frac{\sqrt{10}}{2 a} \ne 0 \mathmr{and} \frac{- b}{2 a} \ne 0$

Let ${x}_{1} = \frac{- b}{2 a} + \frac{\sqrt{10}}{2 a}$

Let ${x}_{2} = \frac{- b}{2 a} - \frac{\sqrt{10}}{2 a}$

Thus ${x}_{1} - {x}_{2} = \cancel{- \frac{b}{2 a}} + \frac{\sqrt{10}}{2 a} \cancel{+ \frac{b}{2 a}} + \frac{\sqrt{10}}{2 a}$

${x}_{1} - {x}_{2} \text{ "=" } \cancel{2} \times \frac{\sqrt{10}}{\cancel{2} a}$

As there is a defined distance between the two points then they do not coincide. Thus there are two separate points.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Lets try out two equations $\text{ } {b}^{2} - 4 a c$ by making up some numbers.

Suppose we had:$\text{ } {b}^{2} - 4 \left(2\right) \left(10\right) = 10$
$\implies {b}^{2} = 10 + 80 = 90 \text{ "->" } b = 3 \sqrt{10}$

Giving:$\text{ } 2 {x}^{2} + 3 \sqrt{10} x + 10$

'.........................................................................
Suppose we had:$\text{ } {b}^{2} - 4 \left(1\right) \left(2\right) = 10$
$\implies {b}^{2} = 10 + 8 \text{ "->" } b = 3 \sqrt{2}$

Giving:$\text{ } 2 {x}^{2} + 3 \sqrt{2} x + 2$
'....................................................................
This is how they look on the graph: