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

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Answer 1 · 2015-05-24T18:26:08

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.

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

:)

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