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

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Answer 1 · 2015-07-23T17:01:36

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

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

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

$2ksqrt6 = 5$

$k = 5/(2sqrt6)$

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

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

$sqrt6p - 5/(2sqrt6) = ± sqrt{k^2+6}$

$sqrt6p = 5/(2sqrt6) ± sqrt{k^2+6}$

$p = 5/(2sqrt6sqrt6) ± sqrt(k^2/6+1}$

$p = 5/12 ± sqrt(1/6 25/24+1}$

$p = 5/12 ± sqrt(25+144}/12$

$p = (5 ± 13}/12$

How do you solve the quadratic equation by completing the square: 6(p^2 – 1) = 5p6(p^2 – 1) = 5p?