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

1 Answer
Oct 16, 2016

Answer:

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

Explanation:

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#