How do you solve #10x^2-31x+15=0# using the quadratic formula?

2 Answers
Jun 12, 2017

Answer:

See a solution process below:

Explanation:

From: http://www.purplemath.com/modules/quadform.htm

The quadratic formula states:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

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

Substituting #10# for #a#; #-31# for #b# and #15# for #c# gives:

#x = (-(-31) +- sqrt((-31)^2 - (4 * 10 * 15)))/(2 * 10)#

#x = (31 +- sqrt(961 - 600))/20#

#x = (31 +- sqrt(361))/20#

#x = (31 +- 19)/20#

#x = 50/20# and #x = 12/20#

#x = 5/2# and #x = 3/5#

Jun 12, 2017

Answer:

#5/2, 6/5#

Explanation:

Use the improved quadratic formula (Google, Socratic Search).
#f(x) = 10x^2 - 31x + 15 = 0#
#D = d^2 = b^2 - 4ac = 961 - 600 = 361# --> #d = +- 19#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 31/20 +- 19/20 = (31 +- 19)/20#
#x1 = 50/20 = 5/2#
#x2 = 12/20 = 3/5#