Question

How do you solve `f(x)=2x^2-12x+17` by completing the square?

Answer 1

`3 +- sqrt(1/2)`

Explanation:

`f(x) = 2x^2 - 12x + 17 => 2(x^2 - 6x) + 17`

so,

`2(x - 3)^2 - 18 + 17`

Hence,

`2x^2 - 12x + 17 = 2(x - 3)^2 - 1`

By 'solving' I assume you mean `2x^2 - 12x + 17 = 0`

i.e. `2(x - 3)^2 - 1 = 0`

`=> (x - 3)^2 = 1/2`

`=> x - 3 = ± sqrt(1/2)`

so, `x = 3 ± sqrt(1/2)`

:)>

Answer 2

`x=3+-sqrt2/2`

Explanation:

To `color(blue)"complete the square"`

add `(1/2" coefficient of x-term")^2`

Require coefficient of `x^2` term to be 1

`f(x)=2(x^2-6x)+17`

`color(white)(f(x))=2(x^2-6xcolor(red)(+9 -9))+17`

Since we have added +9 which is not there we must also subtract 9

`f(x)=2(x-3)^2-18+17`

`rArrf(x)=2(x-3)^2-1`

To solve `color(blue)"equate f(x) to zero"`

`rArr2(x-3)^2-1=0`

`rArr2(x-3)^2=1`

`rArr(x-3)^2=1/2`

`color(blue)"take the square root of both sides"`

`sqrt((x-3)^2)=+-sqrt(1/2)`

`rArrx-3=+-1/sqrt2`

`rArrx=3+-1/sqrt2=3+-sqrt2/2larrcolor(red)" rationalise denominator"`