How do you solve #6x² - 4 = 5x#?

2 Answers
Jul 29, 2015

Answer:

#x=-1/2, 4/3#

Explanation:

#6x^2-4=5x#

Gather all terms on the left side.

#6x^2-5x-4=0#

This is a quadratic equation, #ax^2+bx+c#, where #a=6, b=-5, and c=-4#.

You can use the quadratic formula to solve this equation.

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

#x=(-(-5)+-sqrt(-5^2-4*6*-4))/(2*6)# =

#x=(5+-sqrt(25+96))/12# =

#x=(5+-sqrt(121))/12# =

#x=(5+-11)/12#

Solve for #x#.

#x=(5+11)/12=16/12=4/3#

#x=(5-11)/12=-6/12=-1/2#

#x=-1/2, 4/3#

Jul 30, 2015

Answer:

Solve y = 6x^2 - 5x - 4 = 0 (1)
Ans: - 1/2 and 4/3

Explanation:

I use the new Transforming Method.
Transformed y' = x^2 - 5x - 24. (2). Roots have opposite signs.
Factor pairs of (-24) -> (-2, 12)(-3, 8). This sum is 5 = -b. Two real roots of (2) are: y1 = -3 and y2 = 8.
Back to original equation (1), the 2 real roots are:
#x1 = y1/a = - 3/6 = -1/2# and #x2 = y2/a = 8/6 = 4/3#