How do you complete the square to solve #4x^2 - 7x - 2 = 0#?

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Answer 1 · 2015-05-23T14:37:38

In this way:

#4x^2-7x-2=0rArr4(x^2-7/4x)-2=0rArr#

#4(x^2-7/4x+49/64-49/64)-2=0rArr#

#4(x^2-7/4x+49/64)-49/16-2=0rArr#

#4(x-7/8)^2=49/16+2rArr4(x-7/8)^2=(49+32)/16rArr#

#(x-7/8)^2=81/16*1/4rArr(x-7/8)^2=81/64rArr#

#x-7/8=+-9/8rArrx=7/8+-9/8rArr#

#x_1=7/8-9/8=-2/8=-1/4#

#x_2=7/8+9/8=16/8=2#.