I was just working with a student about a problem that is somewhat similar to this(slightly more complicated). IF YOU HAVE ANY QUESTIONS, LET ME KNOW.. I CAN HELP YOU WITH THIS. Here is THE PROBLEM: x/(x+2) - 7/(5 - x) = 14/(x^2 -3x-10) if you factor the last denominator, you would get (x-5)(x+2), and write the second denominator as -(x-5),.the LCD is a (x-5)(x+2), .simplifying the equation and multiply by the LCD on both sides of the equation, you will get (x-5)*x + (x+2) *7 = 14. Use the distributive property and simplify. You will get x^2 +2x = 0 Factoring this, you get x(x+2) = 0 and you will get x = 0 and x = - 2, BUT x = -2 is an extraneous solution...the only solution you have is x = 0 I hope this helped!
Maybe you should have actually presented an example where there was an extraneous. Otherwise how do students knew what that actually means and when they're going to encounter it?
He did CHECK for extraneous solutions. There just weren't any in the examples he chose. Probably would be more useful to show an example with an extraneous solution.
It’s not inverse operations because the true value of the fractions stay the same when you multiply both sides by the same number. He did it to get the same denominator to subtract the fractions. Inverse operations only applies when you multiply one number (or fraction) to both sides of the equation and the true value changes.
Good video overall, but it would probably be useful to many viewers to show an example that does have an extraneous solution.
I was just working with a student about a problem that is somewhat similar to this(slightly more complicated). IF YOU HAVE ANY QUESTIONS, LET ME KNOW.. I CAN HELP YOU WITH THIS.
Here is THE PROBLEM:
x/(x+2) - 7/(5 - x) = 14/(x^2 -3x-10)
if you factor the last denominator, you would get (x-5)(x+2), and write the second denominator as -(x-5),.the LCD is a (x-5)(x+2), .simplifying the equation and multiply by the LCD on both sides of the equation, you will get (x-5)*x + (x+2) *7 = 14.
Use the distributive property and simplify. You will get x^2 +2x = 0
Factoring this, you get x(x+2) = 0
and you will get x = 0 and x = - 2, BUT x = -2 is an extraneous solution...the only solution you have is x = 0
I hope this helped!
Maybe you should have actually presented an example where there was an extraneous. Otherwise how do students knew what that actually means and when they're going to encounter it?
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hasnt it occured to you to do one WITH an extraneous solution?))
8:59 why not use the quadratic formula (the quadratic equation) ?
This one way of solving quadratic equation called factorising or splitting the middle term.
I hope you got it
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Good job
you didn’t check for any extraneous solutions in this video.
He did CHECK for extraneous solutions. There just weren't any in the examples he chose. Probably would be more useful to show an example with an extraneous solution.
Question on the last example when you multiplied by 7 why didn’t you multiply both sides of the equation because isn’t that inverse operations?
It’s not inverse operations because the true value of the fractions stay the same when you multiply both sides by the same number. He did it to get the same denominator to subtract the fractions. Inverse operations only applies when you multiply one number (or fraction) to both sides of the equation and the true value changes.
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Irrational number please upload
X^2++12==7X
X=3
9++12==7(3)
21=====21
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Thank you
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