How do you solve #4x^2-13x+3=0#?

2 Answers
Jul 15, 2015

Answer:

#x=1/4, 3#

Explanation:

#4x^2-13x+3=0# is a quadratic equation with the form #ax^2+bx+c=0#, where #a=4,# #b=-13,# and #c=3#.

Use the quadratic formula.

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

#x=(-(-13)+-sqrt(-13^2-4*4*3))/(2*4)# =

#x=(13+-sqrt(169-48))/8# =

#x=(13+-sqrt(121))/8# =

#x=(13+-11)/8#

Solve for #x#.

#x=(13+11)/8=24/8=3#

#x=(13-11)/8=2/8=1/4#

Jul 16, 2015

Answer:

Solve: y = 4x^2 - 13x + 3 = 0
Answers: 1/4 and 3

Explanation:

Use the new Transforming Method (Google and Yahoo Search)
Transformed equation: y' = x^2 - 13x + 12. Both roots are positive.
Factor pairs of ac = 12 --> (1, 12) = 13.
y1 = 1 and y2 = 12
x1 = y1/a = 1/4 and x2 = y2/a = 12/4 = 3
Answers: 1/4 and 3.

Note : I advise students to learn and solve quadratic equations by the new Transforming Method. It is simpler and faster than using the formula or the factoring by grouping method.