How do you solve #6x^2-8x-3=0# using the quadratic formula?

2 Answers
Mar 13, 2016

Answer:

#x_(1,2) = (-color(blue)8+- sqrt(color(blue)(64)-72))/(12)= -2/3 +- isqrt(2)/6 = 1/3(-2+-sqrt(2)/2)#
Note this means:
#x_1 = 1/3 (-2+sqrt(2)/2)# and #x_1 = 1/3 (-2-sqrt(2)/2)#

Explanation:

I suggest you commit to memory the "Quadratic Formula" it probably one of the formula that you absolutely positively must know by heart: So here is your Quadratic Formula:
Given a 2nd Order Polynomial,#P_2#:

#P_2 = color(red)ax^2 + color(blue)bx + color(green)c# the Roots or solutions to the equation
#x_1# and #x_2# are given by the Quadratic Formula:

#x_(1,2) = (-color(blue)b+- sqrt(color(blue)(b^2)-4color(red)acolor(green)c))/(2color(red)a)#

Now for your equation: #color(red)6x^2-color(blue)8x-color(green)3=0#

#x_(1,2) = (-color(blue)8+- sqrt(color(blue)(8^2)-4*color(red)6*color(green)3))/(2*color(red)6)#
#x_(1,2) = (-color(blue)8+- sqrt(color(blue)(64)-72))/(12)= -2/3 +- isqrt(2)/6 = 1/3(-2+-sqrt(2)/2)#

Mar 16, 2016

Answer:

#x=(4+-sqrt(34))/6#

Explanation:

#1#. Since the given equation is already in standard form, identify the #color(blue)a,color(darkorange)b,# and #color(violet)c# values. Then plug the values into the quadratic formula to solve for the roots.

#color(blue)6x^2# #color(darkorange)(-8)x# #color(violet)(-3)=0#

#color(blue)(a=6)color(white)(XXXXX)color(darkorange)(b=-8)color(white)(XXXXX)color(violet)(c=-3)#

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

#x=(-(color(darkorange)(-8))+-sqrt((color(darkorange)(-8))^2-4(color(blue)6)(color(violet)(-3))))/(2(color(blue)6))#

#x=(8+-sqrt(64+72))/12#

#x=(8+-sqrt(136))/12#

#x=(8+-2sqrt(34))/12#

#2#. Factor out #2# from the numerator and denominator.

#x=(2(4+-sqrt(34)))/(2(6))#

#x=(color(red)cancelcolor(black)2(4+-sqrt(34)))/(color(red)cancelcolor(black)2(6))#

#color(green)(|bar(ul(color(white)(a/a)x=(4+-sqrt(34))/6color(white)(a/a)|)))#