# How do you find the roots, real and imaginary, of y=4x^2+12 using the quadratic formula?

Oct 16, 2016

$x = \pm 2 \sqrt{3} \textcolor{w h i t e}{. .} i$

#### Explanation:

Given:$\text{ } y = 4 {x}^{2} + 12$

Considered by comparing to $y = {x}^{2}$

As the $4 {x}^{2}$ is positive the general shape is that of $\cup$

The 4 from $4 {x}^{2}$ makes the slope a lot steeper than in $y = {x}^{2}$

As there is no $x$ term (as in $y = a {x}^{2} + \textcolor{red}{b x} + c$ ) the graph is symmetrical about the y-axis.

The +12 'shifts' the graph upwards so that the vertex coincides with the y axis at $y = 12$

$\textcolor{g r e e n}{\text{Thus the solutions for "y=0" are 'imaginary'}}$

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Compare to $y = a {x}^{2} + b x + c$ where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case: $\text{ "a=4"; "b=0"; } c = 12$ giving:

$x = \frac{- 0 \pm \sqrt{{\left(0\right)}^{2} - {\cancel{4}}^{2} \left({\cancel{4}}^{1}\right) \left(12\right)}}{{\cancel{2}}^{1} \left({\cancel{4}}^{1}\right)}$

$x = \pm \sqrt{- 24} \text{ } \to \pm \sqrt{- 1 \times 3 \times {2}^{2}}$

But $\sqrt{- 1} = i$

$x = \pm 2 \sqrt{3} \textcolor{w h i t e}{. .} i$