How do you solve #7x^2 - 12x + 16 = 0 #?

1 Answer
Jan 25, 2017

Answer:

#x = 6/7+-2/7sqrt(19)i#

Explanation:

The given quadratic equation:

#7x^2-12x+16 = 0#

is in the form:

#ax^2+bx+c = 0#

with #a=7#, #b=-12# and #c=16#

This has discriminant #Delta# given by the formula:

#Delta = b^2 - 4ac = (-12)^2-4(7)(16) = 144-448 = -304#

Since #Delta < 0#, this quadratic equation has no Real roots, only Complex one.

We can still find the roots by completing the square.

The difference of squares identity can be written:

#A^2-B^2=(A-B)(A+B)#

We use this with #A=(7x-6)# and #B=2sqrt(19)i# as follows:

#0 = 7(7x^2-12x+16)#

#color(white)(0) = 49x^2-84x+112#

#color(white)(0) = 49x^2-84x+36+76#

#color(white)(0) = (7x)^2-2(7x)(6)+(6)^2+(2sqrt(19))^2#

#color(white)(0) = (7x-6)^2-(2sqrt(19)i)^2#

#color(white)(0) = ((7x-6)-2sqrt(19)i)((7x-6)+2sqrt(19)i)#

#color(white)(0) = (7x-6-2sqrt(19)i)(7x-6+2sqrt(19)i)#

Hence:

#x = 1/7(6+-2sqrt(19)i) = 6/7+-2/7sqrt(19)i#