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

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Answer 1 · 2015-12-31T06:58:03

#x=(1+-isqrt191)/8#

For any quadratic equation #y=ax^2+bx+c#, the roots are found through the formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case, when #y=4x^2-x+12#,

#{(a=4),(b=-1),(c=12):}#

Plug these into the quadratic equation.

#x=(-(-1)+-sqrt((-1)^2-(4xx4xx12)))/(2xx4)#

#x=(1+-sqrt(1-192))/8#

#x=(1+-sqrt(-191))/8#

#x=(1+-isqrt191)/8#

The two solutions of this function are imaginary. The solution has no real solutions and will never cross the #x#-axis.

graph{4x^2-x+12 [-20, 20, -10, 42.6]}