How do you solve f(x)=2x^2-12x+17f(x)=2x2−12x+17 by completing the square?
2 Answers
Explanation:
so,
Hence,
By 'solving' I assume you mean
i.e.
so,
:)>
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(x2−6x)+17
color(white)(f(x))=2(x^2-6xcolor(red)(+9 -9))+17f(x)=2(x2−6x+9−9)+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(x−3)2−18+17
rArrf(x)=2(x-3)^2-1⇒f(x)=2(x−3)2−1 To solve
color(blue)"equate f(x) to zero"equate f(x) to zero
rArr2(x-3)^2-1=0⇒2(x−3)2−1=0
rArr2(x-3)^2=1⇒2(x−3)2=1
rArr(x-3)^2=1/2⇒(x−3)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)√(x−3)2=±√12
rArrx-3=+-1/sqrt2⇒x−3=±1√2
rArrx=3+-1/sqrt2=3+-sqrt2/2larrcolor(red)" rationalise denominator"⇒x=3±1√2=3±√22← rationalise denominator