How do you use the quadratic formula to solve #9x^2-6x-35=0#?
1 Answer
Explanation:
Given:
#9x^2-6x-35 = 0#
Note that this is in standard form:
#ax^2+bx+c = 0#
with
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#
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)#
