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

Question

Eva Adamson · Accepted Answer

#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]}

Eva Adamson · Answer

undefined To find the roots of the quadratic equation $ y = 4x^2 - x + 12 $ using the quadratic formula, you first identify the coefficients $ a $, $ b $, and $ c $ in the equation $ ax^2 + bx + c $.

For the given equation, $ a = 4 $, $ b = -1 $, and $ c = 12 $.

Then, you apply the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Substitute the values of $ a $, $ b $, and $ c $ into the formula:

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

Simplify:

$$ x = \frac{{1 \pm \sqrt{{1 - 192}}}}{{8}} $$

$$ x = \frac{{1 \pm \sqrt{{-191}}}}{{8}} $$

Since the discriminant ($ b^2 - 4ac $) is negative, the roots will be complex (imaginary).

Hence, the roots of $ y = 4x^2 - x + 12 $ are:

$$ x = \frac{1}{8} \pm \frac{\sqrt{191}i}{8} $$

These roots are complex conjugates, where $ i $ is the imaginary unit.