How do you solve the quadratic equation by completing the square: #6(p^2 – 1) = 5p#?

Question

1428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-23T17:01:36

#p \in {3/2, -1}#

#6p^2 - 6 - 5p = 0#

#(sqrt6p - k)^2 = 6p^2 - 2ksqrt6p + k^2#

#2ksqrt6 = 5#

#k = 5/(2sqrt6)#

#(sqrt6p - 5/(2sqrt6))^2 - k^2-6=0#

#(sqrt6p - 5/(2sqrt6))^2 = k^2+6#

#sqrt6p - 5/(2sqrt6) = ± sqrt{k^2+6}#

#sqrt6p = 5/(2sqrt6) ± sqrt{k^2+6}#

#p = 5/(2sqrt6sqrt6) ± sqrt(k^2/6+1}#

#p = 5/12 ± sqrt(1/6 25/24+1}#

#p = 5/12 ± sqrt(25+144}/12#

#p = (5 ± 13}/12#