How do you solve f(x)=2x^2-12x+17f(x)=2x212x+17 by completing the square?

2 Answers

3 +- sqrt(1/2)3±12

Explanation:

f(x) = 2x^2 - 12x + 17 => 2(x^2 - 6x) + 17f(x)=2x212x+172(x26x)+17

so,

2(x - 3)^2 - 18 + 172(x3)218+17

Hence,

2x^2 - 12x + 17 = 2(x - 3)^2 - 1 2x212x+17=2(x3)21

By 'solving' I assume you mean 2x^2 - 12x + 17 = 02x212x+17=0

i.e. 2(x - 3)^2 - 1 = 02(x3)21=0

=> (x - 3)^2 = 1/2(x3)2=12

=> x - 3 = ± sqrt(1/2)x3=±12

so, x = 3 ± sqrt(1/2)x=3±12

:)>

Apr 23, 2017

x=3+-sqrt2/2x=3±22

Explanation:

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

add (1/2" coefficient of x-term")^2(12 coefficient of x-term)2

Require coefficient of x^2x2 term to be 1

f(x)=2(x^2-6x)+17f(x)=2(x26x)+17

color(white)(f(x))=2(x^2-6xcolor(red)(+9 -9))+17f(x)=2(x26x+99)+17

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

f(x)=2(x-3)^2-18+17f(x)=2(x3)218+17

rArrf(x)=2(x-3)^2-1f(x)=2(x3)21

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

rArr2(x-3)^2-1=02(x3)21=0

rArr2(x-3)^2=12(x3)2=1

rArr(x-3)^2=1/2(x3)2=12

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

sqrt((x-3)^2)=+-sqrt(1/2)(x3)2=±12

rArrx-3=+-1/sqrt2x3=±12

rArrx=3+-1/sqrt2=3+-sqrt2/2larrcolor(red)" rationalise denominator"x=3±12=3±22 rationalise denominator