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How to use the discriminant to find out what type of solutions the equation has for #2x^2 + 5x + 5 = 0#?

Answer 1

Eva Abramson

#2x^2+5x+5# has discriminant #Delta = -15#

Since #Delta < 0# the equation has no real solutions, only a pair of complex ones.

#2x^2+5x+5# is of the form #ax^2+bx+c#

with #a=2#, #b=5# and #c=5#

This has discriminant given by the formula:

#Delta = b^2-4ac = 5^2-(4xx2xx5) = 25 - 40 = -15#

Since #Delta < 0# the equation has no real solutions, only complex ones.

The possible cases are:

#Delta > 0# : The equation has two distinct real solutions. If #Delta# is a perfect square (and the coefficients of the quadratic are rational) then the roots are rational.

#Delta = 0# : The equation has one (repeated) real solution.

#Delta < 0# : The equation has two distinct complex roots (which are complex conjugates of one another).

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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.