How do you solve x^2+6x-5=0x2+6x−5=0 by completing the square?

2 Answers
EZ as pi
May 4, 2017

x = 0.742 or x = -6.742x=0.742orx=−6.742

Explanation:

For a quadratic ax^2 +bx+c=0ax2+bx+c=0, there are specific conditions required to be able to write it as the square of a binomial.

a=1, " "(b/2)^2 =ca=1, (b2)2=c

x^2 +6x-5=0x2+6x−5=0

In this case, a=1a=1, but (6/2)^2 = 9 !=-5(62)2=9≠−5

Move -5−5 to the other side and add 99 to both sides.

x^2 +6x color(blue)(+9) = 5color(blue)(+9)x2+6x+9=5+9

(x+3)^2 = 14" "larr(x+3)2=14 ← find the square root of both sides.

x+3 = +-sqrt14x+3=±√14

x = +sqrt14-3" or "x = -sqrt14-3x=+√14−3 or x=−√14−3

x = 0.742 or x = -6.742x=0.742orx=−6.742

Jim G.
May 4, 2017

x=-3+-sqrt14x=−3±√14

Explanation:

"to solve using "color(blue)"completing the square"to solve using completing the square

add (1/2"coefficient of the x-term")^2" to both sides"(12coefficient of the x-term)2 to both sides

"that is " (6/2)^2=9that is (62)2=9

rArr(x^2+6xcolor(red)(+9))-5=0color(red)(+9)⇒(x2+6x+9)−5=0+9

rArr(x+3)^2-5=9⇒(x+3)2−5=9

"add 5 to both sides"add 5 to both sides

(x+3)^2cancel(-5)cancel(+5)=9+5

rArr(x+3)^2=14

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

sqrt((x+3)^2)=color(red)(+-)sqrt14larr" note plus or minus"

rArrx+3=+-sqrt14

"subtract 3 from both sides"

xcancel(+3)cancel(-3)=-3+-sqrt14

rArrx=-3+-sqrt14

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