How do you solve #8x+1=4x^2# by quadratic formula?

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Answer 1 · 2016-02-07T08:17:11

#8x+1=4x^2#

#rarr8x=4x^2-1#

#rarr4x^2-1-8x=0#

In standard form:

#rarr4x^2-8x-1=0#

Now this is a Quadratic equation (in form #ax^2+bx^2+c=0#)

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

In this case #a=4,b=-8,c=-1#

Substitute the values into the formula:

#rarrx=(-(-8)+-sqrt((-8)^2-4(4)(-1)))/(2(4))#

#rarrx=(8+-sqrt(64-(-16)))/8#

#rarrx=(8+-sqrt(64+16))/8#

#rarrx=(8+-sqrt(80))/8#

#rarrx=(8+-sqrt(16*5))/8#

#rarrx=(8+-4sqrt5)/8#

#rarrx=(2+-sqrt5)/2#

Answer 2 · 2016-02-07T08:24:14

#(2+-sqrt(5))/(2)#

quadratic fomula

#ax^2+bx+c=0#

#(-b+-sqrt(b^2 -4ac))/(2a)#

so #8x+1=4x^2#

#rarr4x^2-8x-1=0#

#rarr(-(-8)+-sqrt((-8)^2 -4(4)(-1)))/(2(4))#

#rarr(8+-sqrt(64 +16))/(8)#

#rarr(8+-sqrt(80))/(8)#

#rarr(8+-4sqrt(5))/(8)#

#rarr(2+-sqrt(5))/(2)#