anonymous
  • anonymous
Which quadratic equation is equivalent to (x + 2)2 + 5(x + 2) – 6 = 0? (u + 2)^2 + 5(u + 2) – 6 = 0 where u = (x – 2) u^2 + 4 + 5u – 6 = 0 where u = (x – 2) u^2 + 5u – 6 = 0 where u = (x + 2) u^2 + u – 6 = 0 where u = (x + 2)
Mathematics
jamiebookeater
  • jamiebookeater
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misty1212
  • misty1212
lol HI again!
anonymous
  • anonymous
hey girl!
misty1212
  • misty1212
\[ (x + 2)^2 + 5(x + 2) – 6 = 0\] put \(\color{red}u=x+2\) get \[\huge\color{red}u^2+5\color{red}u-6=0\]

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anonymous
  • anonymous
or just go with C since it is always C
misty1212
  • misty1212
lol ikr!
anonymous
  • anonymous
When in doubt pick C lol.
anonymous
  • anonymous
Which equation is quadratic in form? 4(x – 2)2 + 3x – 2 + 1 = 0 8x^5 + 4x^3 + 1 = 0 10x^8 + 7x^4 + 1 = 0 9x^16 + 6x^4 + 1 = 0
anonymous
  • anonymous
misty1212
  • misty1212
quadratic means hmm this is a trick
misty1212
  • misty1212
i think they want you to say \(10x^8 + 7x^4 + 1 = 0\) is in "quadratic form" because you can make \(u=x^4\) and it becomes \[10u^2+7u+1=0\]
misty1212
  • misty1212
the first one is in fact a quadratic equation, but C (as usual C) is in "quadratic form" go with that one
anonymous
  • anonymous
What substitution should be used to rewrite 4x^12 – 5x6 – 14 = 0 as a quadratic equation? u = x^2 u = x^3 u = x^6 u = x^12
anonymous
  • anonymous
misty1212
  • misty1212
the middle term is has \(x^6\) so you can use \(u=x^6\)
misty1212
  • misty1212
ooh what a surprise, C again...
anonymous
  • anonymous
Thank you! What substitution should be used to rewrite 16(x^3 + 1)^2 – 22(x^3 + 1) – 3 = 0 as a quadratic equation? u = (x^3) u = (x^3 + 1) u = (x^3 + 1)^2 u = (x^3 + 1)^3
misty1212
  • misty1212
wow not C
misty1212
  • misty1212
you can use \(u=(x^3+1)\)
anonymous
  • anonymous
What are the solutions of the equation x^4 + 3x^2 + 2 = 0? Use u substitution to solve.
misty1212
  • misty1212
\[ x^4 + 3x^2 + 2 = 0\] put \[u=x^2\] get \[y=u^2+3u+2=0\] or \[(u+1)(u+2)=0\] but there will be no real solution only complex ones
misty1212
  • misty1212
is there a typo in the question?
anonymous
  • anonymous
A.) ±i (sqrt)2 and x = ±1 B) ±i (sqrt)2 and x = ±i C)±(sqrt)2 and x = ±i D)± (sqrt)2 and x = ±1
misty1212
  • misty1212
oh they allow complex solutions ok
anonymous
  • anonymous
yes it just took me a minute to type out! :]
misty1212
  • misty1212
\[u+1=00\\ u=-1\\ ^2=-1\\ x=\pm i\] is one pair
misty1212
  • misty1212
the other is \(x=\pm\sqrt{2}i\)
misty1212
  • misty1212
looks like they put the \(i\) first , go with B
anonymous
  • anonymous
What are the solutions of the equation x^4 – 5x^2 – 14 = 0? Use factoring to solve. Same as before.
misty1212
  • misty1212
yeah same as before \[u=x^2\] get \[u^2-5u-14=0\] or \[(u-7)(u+2)=0\] so \[u=7\]or \[u=-2\]
misty1212
  • misty1212
then \[x^2=7\\ x=\pm\sqrt7\] or \[x^2=-2\\ x=\pm\sqrt{2}i\]
anonymous
  • anonymous
yayy thats what i had :)
anonymous
  • anonymous
thank you!
misty1212
  • misty1212
one to go?
misty1212
  • misty1212
or was that it?

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