How does #t^2-2t-1=0# become #t=1+-sqrt2#?

2 Answers
Jim G.
May 29, 2018

#"see explanation"#

Explanation:

#"solve for t using the method of "color(blue)"completing the square"#

#• " the coefficient of the "t^2" term must be 1 which it is"#

#• " add "(1/2"coefficient of the t- term ")^2" to both sides"#

#"add 1 to both sides"#

#t^2-2t=1#

#t^2+2(-1)t color(red)(+1)=1color(red)(+1)#

#(t-1)^2=2#

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

#sqrt((t-1)^2)=+-sqrt2larrcolor(blue)"note plus or minus"#

#t-1=+-sqrt2#

#"add 1 to both sides to obtain"#

#t=1+-sqrt2#

MathFact-orials.blogspot.com
May 30, 2018

An alternate way using the quadratic formula:

Explanation:

Another way to find solutions for #t# is to use the quadratic formula:

# t = (-b \pm sqrt(b^2-4ac)) / (2a) #

with #a=1, b=-2, c=-1#

# t = (2 \pm sqrt((-2)^2-4(1)(-1))) / (2(1)) #

# t = (2 \pm sqrt(4+4)) / 2 #

# t = (2 \pm sqrt8) / 2 #

# t = (2 \pm 2sqrt2) / 2 #

# t = 1 \pm sqrt2 #

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