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How do you find all the zeros of #y=x^2-11x+30# with its multiplicities?

Answer 1

Caleb Alexander

#x = 5, 6#

Write out a set of brackets like this

#(x + a)(x + b)#

You should find two constants, #a# and #b#, such that

#a + b = -11# #a * b = 30#

Which you can do by trial and error, so

#a = -5# #b = -6#

Therefore the fully factorised form of the equation is

#(x - 5)(x - 6) = 0#

Solving for the roots or zeros of this:

For the entire quadratic to equal zero, one or both of the pairs of brackets must equal #0#, so solve for each individual one and you will end up with two answers.

#x - 5 = 0 -> x = 5# #x - 6 = 0 -> x = 6#

Therefore,

#x = 5, 6#

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Answer 2

Benjamin Achtenberg

To find all the zeros of (y = x^2 - 11x + 30) with their multiplicities, you can factor the quadratic expression or use the quadratic formula. Factoring the expression gives (y = (x - 5)(x - 6)), which shows that the zeros are (x = 5) and (x = 6). Both zeros have a multiplicity of 1 because they are distinct.

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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.