Question 809394
<pre>
There are two ways, by completing the square:

First way:

    y = 2x² - x - 1

Get the constant term on the left by adding 1 to both sides:

y + 1 = 2x² - x

To get the first term on the right to just x² by multiplying
every term through by {{{1/2}}}

{{{1/2}}}y + {{{1/2}}} = x² - {{{1/2}}}x

Complete the square on the right side of the equation:
1.  Multiply the coefficient of x which is {{{-1/2}}} by {{{1/2}}}, getting {{{-1/4}}}
2.  Square that amount {{{(1/4)^2}}} = {{{1/16}}}

3. Add that amount {{{1/16}}} to both sides of the equation:

{{{1/2}}}y + {{{1/2}}} + {{{1/16}}}= x² - {{{1/2}}}x + {{{1/16}}}


Factor the right side as a perfect square:

{{{1/2}}}y + {{{1/2}}} + {{{1/16}}}= (x - {{{1/4}}})²

Clear of fractions by multiplying every term on both sides by 16

8y + 8 + 1 = 16(x - {{{1/4}}})²

Combine 8 + 1 as 9

8y + 9 = 16(x - {{{1/4}}})

Subtract 9 from both sides

8y = 16(x - {{{1/4}}})² - 9

Solve for y by dividing every term by 8

 y = 2(x - {{{1/4}}})² - {{{9/8}}}

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Second way:

    y = 2x² - x - 1

Factor 2 out of the first two terms:

    y = 2(x² - {{{1/2}}}x) - 1

Change the parentheses to brackets so they can hold parentheses:

    y = 2[x² - {{{1/2}}}x] - 1

Complete the square in the parentheses:
1.  Multiply the coefficient of x which is {{{-1/2}}} by {{{1/2}}}, getting {{{-1/4}}}
2.  Square that amount {{{(1/4)^2}}} = {{{1/16}}}

3. Add and subtract that amount {{{1/16}}} inside the brackets:

    y = 2[x² - {{{1/2}}}x + {{{1/16}}} - {{{1/16}}}] - 1

Factor the first three terms in the bracket as a perfect square:

    y = 2[(x - {{{1/4}}})² - {{{1/16}}}] - 1

Remove the bracket by distributing the 2, leaving the parentheses intact:

    y = 2(x - {{{1/4}}})² - {{{2*expr(1/16)}}} - 1

    y = 2(x - {{{1/4}}})² - {{{1/8}}} - 1

Combine the last two terms, by writing 1 as {{{8/8}}}

    y = 2(x - {{{1/4}}})² - {{{1/8}}} - {{{8/8}}}

    y = 2(x - {{{1/4}}})² - {{{9/8}}}

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Take your pick.

Edwin</pre>