why?!! How to know we've solved the equation?
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What are your thoughts?
Please share your work with us ...even if you know it is wrong
If you are stuck at the beginning tell us and we'll start with the definitions.
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Bob Brown MSEE
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Example #1 "Good"
(x-4) (x-5) = 0 is true if
either (x-4)=0 or (x-5)=0
are true for different x values.
Example #2 "Not Good"
(x-4) (x-5) = 1
then BOTH (x-4) = 1 AND (x-5) = 1
Need to be true for the SAME x value. (impossible)
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When you get down to (variable) equals (a number), you've found the solution, so the equation is solved.
Please show your work for what you're proposing should be an acceptable answer. You started by multiplying out the left-hand side, and subtracting the 1 over to the left-hand side, so that you can then use the "zero product property", and... then what?
Please be complete. Thank you!
thnx ...but why?!
But why equality in each of the brackets to 0 Derives we found all the solutions?
:smile:
But why equality in each of the brackets to 0 Derives we found all the solutions?
I'm afraid I don't understand what you're asking. You'll always find all of the solutions. Sometimes there's only one value of x that works, and sometimes there's more than one, or in some cases, no value of x will work at all and there are no solutions. If you're confused about why the process is different when the equation equals 0, then I can try and explain that.
The first of your two specific examples is (x-4)(x-5)=0. Here, you have two expressions which multiply to zero. So, if (something) times (something else) equals 0, can you see that at least one of them must be 0?
Your second example says (x-4)(x-5)=1. We can't use the same strategy as before, because now we have (something) times (something else) equals 1. If we apply the same logic and say that one of them must be 1... well, we know that can't work because \(\displaystyle 2 \cdot 1\ne 1\), \(\displaystyle 3 \cdot 1\ne 1\), etc. Stapel gave you a good hint for how to start this one. Did you follow those steps? What did you get?
The first of your two specific examples is (x-4)(x-5)=0. Here, you have two expressions which multiply to zero. So, if (something) times (something else) equals 0, can you see that at least one of them must be 0?
Your second example says (x-4)(x-5)=1. We can't use the same strategy as before, because now we have (something) times (something else) equals 1. If we apply the same logic and say that one of them must be 1... well, we know that can't work because \(\displaystyle 2 \cdot 1\ne 1\), \(\displaystyle 3 \cdot 1\ne 1\), etc. Stapel gave you a good hint for how to start this one. Did you follow those steps? What did you get?
(x-4)(x-5)=0
x=4 x=5
Why do students need to believe that all the solutions? not more!
(x-2)(x-3)(x-5)(x+10)=0
x=2 x=3 x=5 x=-10
Why do students need to believe that all the solutions? not more!
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The "(these factors) equal (zero)" form of an equation uses the fact that, if a product (like "xyz") equals zero, then at least one of the factors (x, or y, or z) must equal zero (x = 0, or y = 0, or z = 0). This is the ONLY way to get an answer of zero. And this is why we have to have factored form; it's the only way we get to use the "must equal zero" property.
We also know that, graphically, "(factors) equal (zero)" is the same as "(factored or otherwise) crosses the line (y = 0, or the x-axis)". This is how we can connect the factored form to the graph. If we can't factor, we have to use other methods (like numerical approximation) to figure out the zeroes (or solutions). But x-intercepts are always solutions of "(whatever function) equals (zero)".
We can do "(whatever function) equals (some other number)", but zero is generally easier, is certainly nicer to graph, and is the only form that works when our (whatever function) happens to factor. So I guess the question becomes:
When the function (set equal to zero) is factored and the zeroes are found, what other solutions would you propose might exist? :shock: