How do you find the zeros of #g(x) = 4x^2-2x-3# using the quadratic formula?

1 Answer
Oct 26, 2016

Answer:

#x = 1/4+-sqrt(13)/4#

Explanation:

#g(x) = 4x^2-2x-3#

is in the standard form #ax^2+bx+c#, with #a=4#, #b=-2# and #c=-3#

We can find the zeros of #g(x)# using the quadratic formula:

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

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

#color(white)(x) = (2+-sqrt(4+48))/8#

#color(white)(x) = (2+-sqrt(52))/8#

#color(white)(x) = (2+-sqrt(2^2*13))/8#

#color(white)(x) = (2+-2sqrt(13))/8#

#color(white)(x) = 1/4+-sqrt(13)/4#

#color(white)()#
Footnotes

Given any quadratic equation in the form #ax^2+bx+c = 0#,

the quadratic formula:

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

is very useful, but do you know where it comes from, how it works or how to derive it yourself?

Here's one way...

Given #ax^2+bx+c = 0#, (with #a != 0#) we have:

#0 = ax^2+bx+c#

#color(white)(0) = a(x^2+2b/(2a)x + b^2/(4a^2) - b^2/(4a^2) + c/a)#

#color(white)(0) = a((x+b/(2a))^2-(b^2-4ac)/(4a^2))#

#color(white)(0) = a((x+b/(2a))^2-(sqrt(b^2-4ac)/(2a))^2)#

#color(white)(0) = a((x+b/(2a))-(sqrt(b^2-4ac)/(2a)))((x+b/(2a))+(sqrt(b^2-4ac)/(2a)))#

#color(white)(0) = a(x+(b-sqrt(b^2-4ac))/(2a))(x+(b+sqrt(b^2-4ac))/(2a))#

Hence:

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