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

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Answer 1 · 2016-10-16T09:57:05

$x=+-2sqrt(3)color(white)(..)i$

Given: $" "y=4x^2+12$

Considered by comparing to $y=x^2$

As the $4x^2$ is positive the general shape is that of $uu$

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

As there is no $x$ term (as in $y=ax^2+color(red)(bx)+c$ ) the graph is symmetrical about the y-axis.

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

$color(green)("Thus the solutions for "y=0" are 'imaginary'")$

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Compare to $y=ax^2+bx+c$ where $x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case: $" "a=4"; "b=0"; "c=12$ giving:

$x=(-0+-sqrt((0)^2-cancel(4)^2(cancel(4)^1)(12)))/(cancel(2)^1(cancel(4)^1))$

$x=+-sqrt(-24)" "->+-sqrt(-1xx3xx2^2)$

But $sqrt(-1)=i$

$x=+-2sqrt(3)color(white)(..)i$

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