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How do you solve #x^2 - 2x - 5 = 0#?

Answer 1

Eva Adamson

The solutions are:
#color(blue)(=1+sqrt(6)#

#color(blue)(=1-sqrt(6)#

# x^2−2x−5=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where: #a=1, b=-2, c= -5#

The Discriminant is given by: #Delta=b^2-4*a*c#

# = (-2)^2-(4*1* (-5))#

# = 4 +20 = 24#

The solutions are found using the formula #x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-2)+-sqrt(24))/(2*1) = (2+-2sqrt(6))/2#

#=(cancel2(1+-sqrt(6)))/cancel2#

The solutions are: #color(blue)(=1+sqrt(6)#

#color(blue)(=1-sqrt(6)#

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Answer 2

Ethan Adkins

To solve the equation x^2 - 2x - 5 = 0, you can use the quadratic formula:

x = (-(-2) ± √((-2)^2 - 41(-5))) / (2*1)

Simplify the formula:

x = (2 ± √(4 + 20)) / 2

x = (2 ± √24) / 2

x = (2 ± 2√6) / 2

x = 1 ± √6

So, the solutions to the equation are:

x = 1 + √6 and x = 1 - √6.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.