How do you use the quadratic formula to solve #9x^2-6x-35=0#?

1 Answer
May 7, 2017

Answer:

#x = 7/3" "# or #" "x = -5/3#

Explanation:

Given:

#9x^2-6x-35 = 0#

Note that this is in standard form:

#ax^2+bx+c = 0#

with #a=9#, #b=-6# and #c=-35#

The roots are given by the quadratic formula:

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

#color(white)(x) = (-(color(blue)(-6))+-sqrt((color(blue)(-6))^2-4(color(blue)(9))(color(blue)(-35))))/(2(color(blue)(9)))#

#color(white)(x) = (6+-sqrt(36+1260))/18#

#color(white)(x) = (6+-sqrt(1296))/18#

#color(white)(x) = (6+-36)/18#

#color(white)(x) = (1+-6)/3#

That is:

#x = 7/3" "# or #" "x = -5/3#

#color(white)()#
Notes

The quadratic formula is very useful and worth memorizing, but I would strongly recommend that you learn how to derive it from scratch - if you do not know already.

Here's one way, making use of the difference of squares identity:

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

Given:

#ax^2+bx+c = 0#

We can write:

#0 = ax^2+bx+c#

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

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