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

2 Answers
Meave60
Jul 15, 2015

#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#

Nghi N.
Jul 16, 2015

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.

Related questions
See all questions in Comparing Methods for Solving Quadratics
Impact of this question
7841 views around the world
You can reuse this answer
Creative Commons License