How do you solve the quadratic #8b^2-1=-78# using any method?

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Answer 1 · 2017-01-02T12:40:05

#b=(isqrt(77)sqrt(2))/4=(isqrt(154))/4 ~~i3.1024...#

Add 1 to both sides

#8b^2=-77#

Divide both sides by 8

#b^2=-77/8 -9.625 larr" as a check value"#

#b=sqrt(-77)/sqrt(2xx2^2)#

#b=(isqrt(77))/(2sqrt(2))xx1=(isqrt(77))/(2sqrt(2))xxsqrt(2)/sqrt(2)#

#b=(isqrt(77)sqrt(2))/4=(isqrt(154))/4 ~~i3.1024...#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check

#b^2=((isqrt(154))/4)^2#

#b^2=-154/16#

#b^2=-9.625 larr" check confirmed"#