How do you solve using the completing the square method x(6x−5)=6?

1 Answer
Tony B
May 9, 2016

x=+32 and −23

Explanation:

Multiply out the racket giving: 6x2−5x=6

Subtract 6 from both sides

6x2−5x−6=0

Write as 6(x2−56x)−6=0

Take the square outside the bracket and add the correction constant k

6(x−56x)2−6+k=0

Remove the x from −56x

6(x−56)2−6+k=0

Multiply the −56 by (12)

6(x−512)2−6+k=0

'~~~~~~~~~~~~~ Comment ~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the part of (x−512)(x−512)

The (−512)2 is an introduced value that is not in the original equation so we remove it by subtraction. However, this introduced error is in fact 6(−512)2 due to the 6 outside the brackets
⇒k=(−1)×6×(−512)2=−1124=−2524
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

6(x−512)2−6+k=0 → 6(x−512)2−6−2524=0

⇒6(x−512)2−16924=0

6(x−512)2=16924

(x−512)2=169144

Taking the square root of both sides

x−512=±√169√144=±1312

x=512±1312

x=+1812 and −812

x=+32 and −23

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