How do you solve #(x-8)(x-1)=0#?

Answer 1

One way of seeing this is using some logic and math.

If you have a product that equals zero, as in this case, you must agree that at least one of the factors is zero.

Thus, either #x-8# and/or #x-1# equals zero.
Thus, #x_1=8# and #x_2=1#.

Another way is distributing these factors. To distribute, you need to multiply each term of the first by all the terms in the second, as follows:

#(x-8)(x-1)#
#(x*x)+(x*(-1))+((-8)*x)+((-8)(-1))# #x^2-x-8x+8# #x^2-9x+8#

And now let's find its roots. I'll use Bhaskara here.

#(9+-sqrt(81-4(1)(8)))/2# #(9+-7)/2# #x_1=(9+7)/2=8# #x_2=(9-7)/2=1#

:)

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Answer 2

To solve the equation ((x-8)(x-1) = 0), you set each factor to zero and solve for (x):

  1. Set (x-8 = 0) and solve for (x): (x - 8 = 0) (x = 8)

  2. Set (x-1 = 0) and solve for (x): (x - 1 = 0) (x = 1)

So, the solutions to the equation are (x = 8) and (x = 1).

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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.

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.

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.

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.

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