How do you solve #4x ^ { 2} - 5= - 3+ 7x#?

1 Answer
smendyka
Nov 17, 2016

The solutions are: #x = -1/4# and #x = 2#

Explanation:

First create a quadratic by moving all the terms to the left side of the equation:

#4x^2 - 5 + 3 - 7x = -3 + 7x + 3 - 7x#

#4x^2 - 7x - 2 = 0#

Now after some playing with terms, specifically pairs of (1,4) and (2,2) for the coefficients of #x# and other pairs for the constants the quadratic can be factored to:

#(4x + 1)(x - 2) = 0#

Solving for #(4x + 1)# gives:

#((4x + 1)(x - 2))/(x - 2) = 0/(x-2)#

#4x + 1 = 0#

#4x + 1 - 1 = 0 - 1#

#4x = -1#

#4x/4 = -1/4#

#x = -1/4#

And solving for (x - 2) gives:

#((4x + 1)(x - 2))/(4x + 1) 0/(4x + 1)#

#x - 2 = 0#

#x - 2 + 2 = 0 + 2#

#x = 2#

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