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How do you solve #5x^2 - 2x – 2 = 0#?

Answer 1

Kennedy Adams

The solutions are:

#color(blue)(x= (1+sqrt11)/5#

#color(blue)(x= (1-sqrt11)/5#

#5x^2−2x – 2=0#

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

The Discriminant is given by:

#color(blue)(Delta=b^2-4*a*c#

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

#=4+40#

# = 44#

The solutions are found using the formula:

#color(blue)(x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-2)+-sqrt(44))/(2*5) = (2+-2sqrt(11))/10#

#=(cancel2(1+-sqrt11))/cancel10 = (1+-sqrt11)/5#

The solutions are:

#color(blue)(x= (1+sqrt11)/5#

#color(blue)(x= (1-sqrt11)/5#

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Answer 2

Eva Adamson

To solve the quadratic equation 5x^2 - 2x - 2 = 0, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Here, a = 5, b = -2, and c = -2. Substituting these values into the quadratic formula gives you: x = [(-(-2) ± √((-2)^2 - 4(5)(-2))) / (2(5))] = [ (2 ± √(4 + 40)) / 10 ] = [ (2 ± √44) / 10 ]. So, the solutions are x = (2 + √44)/10 and x = (2 - √44)/10, which can also be simplified if required.

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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.