How do you solve #2x^2 - 3x + 4 = 0# using the quadratic formula?

3 Answers
Jan 20, 2017

Answer:

#x = 3 pm sqrt(7)/(2sqrt(2))#

Explanation:

#f(x) = (xsqrt(2) -3/sqrt(2))^2 + 4 - 9/4 = 0#
if #xsqrt(2) -3 /sqrt(2) = pm sqrt(7)/4#
or #x = 3 pm sqrt(7)/(2sqrt(2))#

Jan 20, 2017

Answer:

#x = 3/4+-sqrt(23)/4i#

Explanation:

Given:

#2x^2-3x+4 = 0#

Note that this is of the form:

#ax^2+bx+c = 0#

with #a=2#, #b=-3# and #c=4#

The discriminant #Delta# of this quadratic is given by the formula:

#Delta = b^2-4ac = (-3)^2-4(2)(4) = 9-32 = -23#

Since #Delta < 0# we can tell that this quadratic has no Real zeros - only Complex ones.

We can still use the quadratic formula to find them:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (3+-sqrt(-23))/4#

#color(white)(x) = (3+-sqrt(23)i)/4#

#color(white)(x) = 3/4+-sqrt(23)/4i#

Jan 20, 2017

#2x^2-3x+4 = 0#

The solutions to #ax^2+bx+c=0# are given by

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

In this case, we get

#x = (-(-3)+-sqrt((-3)^3-4(2)(4)))/(2(2))#

# = (3 +- sqrt(9-32))/4#

# = (3 +- sqrt (-23))/4#

The solutions are imaginary

#3/4 +- sqrt23/4 i#