anonymous
  • anonymous
What two numbers multiply to 196? If they do not add to 27, how do I create an equation that results in at least one extraneous solution?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
What ?
anonymous
  • anonymous
haha, I'm trying by myself to come up with an equation that has an extraneous solution. I've gotten to the point where I need to figure out what two numbers multiply to 196, but add to 27.
anonymous
  • anonymous
But if there aren't any numbers that can do that, then I don't know how to make an equation with an extraneous solution. :/

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anonymous
  • anonymous
ammm....
anonymous
  • anonymous
?
anonymous
  • anonymous
can u write your means ???
anonymous
  • anonymous
Here's an example of what I'm trying to do
1 Attachment
anonymous
  • anonymous
I need to come up with an example like that one.
phi
  • phi
you would get an extraneous solution if you need to take the negative root for example \[ \sqrt{x^2} = - 2 \] in these problems we normally only allow the positive value for the square root
anonymous
  • anonymous
So I would just have the root equal a negative number?
phi
  • phi
notice in your example they have \[ \sqrt{1} - 7 =? -8 \\ 1 - 7 =? -8 \\ -6 ≠ -8\] but if we took the negative square root , because you might think - is allowed in \( ±\sqrt{x} \) you would get \[ \sqrt{1} - 7 =? -8 \\ -1 - 7 =-8 \]
anonymous
  • anonymous
Ok, I think I understand where you are coming from. But how am I going to come up with an equation using that information?
phi
  • phi
we will get into trouble if we have \[ \sqrt{x+a}=−x \] because the left side will be positive (we only take the + square root) but -x will be negative (assuming x is positive) to find a "problem equation" we can work backwards: square both sides x+a = x^2 or x^2 - x - a = 0 we want a quadratic with an x= positive and x= negative (so that the positive x will be extraneous (see above) for example (x -2)(x+1) = 0 will have solution x= 2 and x= -1 expand to get x^2 -x -2 = 0 add x+2 to both sides x^2 = x+2 take the square root \[ x=\sqrt{x+2}\] I think this will have an extraneous solution
phi
  • phi
fixed a typo

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