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

2 Answers

Answer:

#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)#

:)>

Apr 23, 2017

Answer:

#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"#