How do you solve #8x^2-6x+1=0#?

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Answer 1 · 2017-03-30T20:33:05

#x=1/4 or x=1/2#

Factorise the quadratic trinomial and then let each factor be equal to 0.

#8x^2-6x +1 =0#

(Find factors of 8 which add up 6.)

#(4x-1)(2x-1)=0#

If #4x-1=0" " rarr 4x=1" "rarr x =1/4#

If #2x-1=0" "rarr2x=1" "rarrx=1/2#

Answer 2 · 2017-03-31T21:05:20

# x= 1/2 # or #x = 1/4# (You usually solve equations of this type using the discriminant)

In any equation with the form of: # ax^2 + bx + c = 0 # you can calculate the discriminant ( D or Δ) by using the formula:

# D = b^2 - 4 * ac #

Then, there are three different cases of finding the roots depending on the discriminant

No1: if D < 0 then the equation has no rootes

No2: if D = 0 then the equation has one answer which you can find using
# x = - ( b / (2*a)) #

No3: if D>0 then the equation has two roots which you can find using

# x = ( -b +- sqrtD ) / ( 2 * a ) #

Let's take your example:
#8x^2 - 6x + 1 = 0 #

#D= 36 - 32 = 4 D>0#

#x1= (6 + 2) / 16 #
#x1= 1 / 2 #

And
#x2= ( 6 - 2) / 16#
# x2= 1 / 4 #