How do you solve #5x ^ { 2} - 17x + 6= 0#?

2 Answers
Jun 14, 2017

Answer:

Break the trinomial down into two binomials and solve for each binomial

Explanation:

# Ax^2 + Bx + C# is the standard form of the trinomial.

The B in this equation is negative.
The C in this equation is positive.

This means that for B to be negative both factors of C must be negative.

The sum of the products of the A factors times the C factors, must equal -17.

# (5 xx 3 = 15 ) + ( 1 xx 2 = 2) #

# -15 + -2 = -17# so

# ( 5x - 2) xx ( 1x -3) = 5x^2 -17x + 6# so

# 5x -2 = 0 # add 2 to both sides

# 5x -2 +2 = 0 + 2 # this gives

# 5x = 2 # divide both sides by 5

# 5x/5 = 2/5 # this gives.

# x = 2/5 #

# 1x - 3 = 0# add three to both sides.

# 1x -3 +3 = 0 + 3 # this gives.

# 1x = 3 #

So x equals both 2/5 and 3

Jun 30, 2017

Answer:

2/5 and 5

Explanation:

#y = 5x^2 - 17x + 6 = 0#
Use the new Transforming Method (Socratic, Google Search)
Transformed equation:
#y' = x^2 - 17x + 30 = 0#
Method: find 2 real roots of y', then, divide them by a = 5.
Find 2 real roots of y', knowing sum (-b = 17) and product (ac = 30).
They are : 2 and 15.
Back to original y, the 2 real roots are:
#x1 = 2/a = 2/5#, and
#x2 = 15/a = 15/5 = 3#