How do you solve #4+4x=2x^2# using the quadratic formula?

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Answer 1 · 2017-05-18T15:42:58

#x=1+-sqrt(3)" "#Exact value

#x~~-0.732050....-> -0.73# to 2 decimal places

#x~~color(white)(-)2.732050...... ->+2.73 # to 2 decimal places

Subtract 4 and #4x# from both sides giving:

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

Consider the standard form of #y=ax^2+bx+c#

That at #y=0# we have #x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case: #a=2"; "b=-4"; "c=-4#

so by substitution we have:

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

#x=1+-sqrt(48)/4#

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Note that #sqrt(48)/4# is the same as #sqrt(48/4^2)=sqrt(3)#

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#x=1+-sqrt(3)" "#Exact value

#x~~-0.732050....-> -0.73# to 2 decimal places

#x~~color(white)(-)2.732050...... ->+2.73 # to 2 decimal places