How do you solve #7x^2 +28=53x#?

2 Answers
Mar 13, 2016

Answer:

Factor by grouping to find zeros:

#x = 4/7# and #x=7#

Explanation:

For subtract #53x# from both sides to get:

#7x^2-53x+28 = 0#

We can split the middle term and factor by grouping as follows:

#0 =7x^2-53x+28#

#=7x^2-49x-4x+28#

#=(7x^2-49x)-(4x-28)#

#=7x(x-7)-4(x-7)#

#=(7x-4)(x-7)#

Hence zeros #x=4/7# and #x=7#

Mar 13, 2016

Answer:

(4/7) and 7.

Explanation:

I use the new Transforming Method (Google, Yahoo Search).
#y = 7x^2 - 53x + 28 = 0#
Transformed equation: #y' = x^2 - 53x + 196#
Factor pairs of (ac = 196) --> (2, 98)(4, 49). This sum is 53 = -b. Then the 2 real roots of y' are: 4 and 49.
Back to original equation y, the 2 real roots are: #(4/7)# and
#(49/7 = 7)#

NOTE . There is no need to factor by grouping and solving the 2 binomials.