# How do you find the discriminant and how many and what type of solutions does x^2+3x+7=0 have?

May 10, 2015

Let's get some theoretical background first.
Assume you have to solve a general quadratic equation
$a {x}^{2} + b x + c = 0$
You have no recollection of the formula for its solution (like most people). How should you solve it?
Let's apply the logic.

If your equation looks like this
$\alpha {\left(x + \beta\right)}^{2} + \gamma = 0$,
the solution is easy:

1. You subtract $\gamma$ from both sides, getting
$\alpha {\left(x + \beta\right)}^{2} = - \gamma$
2. Divide by $\alpha$ both sides, getting
#(x+beta)^2=-gamma/alpha
3. Extract a square root from both side using positive and negative values of a square root, getting
$x + \beta = \sqrt{- \frac{\gamma}{\alpha}}$
4. Subtract $\beta$ from both sides, getting
${x}_{1 , 2} = \pm \sqrt{- \frac{\gamma}{\alpha}} - \beta$

Our general equation is not of the form we can easily solve, as above, but can be transformed into this form.
Let's equate two forms and find the necessary $\alpha , \beta , \gamma$ coefficients:
$a {x}^{2} + b x + c = \alpha {\left(x + \beta\right)}^{2} + \gamma$

Open the parenthesis on the right:
$a {x}^{2} + b x + c = \alpha {x}^{2} + 2 \alpha \beta x + \alpha {\beta}^{2} + \gamma$

Now you can determine $\alpha , \beta , \gamma$ coefficients by comparing the coefficient at $x$ in different degrees:
$a = \alpha$ (coefficient at ${x}^{2}$)
$b = 2 \alpha \beta$ (coefficient at $x$)
$c = \alpha {\beta}^{2} + \gamma$ (free coefficient)
from which follows:
$\alpha = a$
$\beta = \frac{b}{2 \alpha} = \frac{b}{2 a}$
$\gamma = c - \alpha {\beta}^{2} = c - a {b}^{2} / \left(4 {a}^{2}\right) = c - {b}^{2} / \left(4 a\right) = \frac{4 a c - {b}^{2}}{4 a}$

Now we use the formula that expressed the solution in terms of $\alpha , \beta , \gamma$, substituting these coefficients with their representation in terms of $a , b , c$.

${x}_{1 , 2} = \pm \sqrt{- \frac{\gamma}{\alpha}} - \beta = + = \sqrt{\frac{- 4 a c + {b}^{2}}{4 a}} - \frac{b}{2 a} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The most involved part of this expression is under the square root, it's called discriminant of this quadratic equation:
$D = {b}^{2} - 4 a c$

If $D = {b}^{2} - 4 a c > 0$, there are two distinct real solution to this quadratic equation because the square root of the positive number has two distinct real values.

If $D = {b}^{2} - 4 a c = 0$, there is only one real solution since the square root equals to 0 and our $\pm$ operation renders always $0$.

Finally, if $D = {b}^{2} - 4 a c < 0$, there are no real solution because there are no real values for a square root of a negative number.

In a particular case of this problem with an equation
${x}^{2} + 3 x + 7 = 0$
the coefficients are:
$a = 1 , b = 3 , c = 7$
The discriminant $D = {b}^{2} - 4 a c = {3}^{2} - 4 \cdot 1 \cdot 7 = - 19$
It's negative. Therefore, there are no real solutions.

A more detailed explanation of topics touched in this answer can be found on Unizor by following the menu items Algebra - Quadratic Equations.