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)

- anonymous

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- misty1212

lol HI again!

- anonymous

hey girl!

- 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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## More answers

- anonymous

or just go with C since it is always C

- misty1212

lol ikr!

- anonymous

When in doubt pick C lol.

- 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

@misty1212

- misty1212

quadratic means hmm this is a trick

- 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

the first one is in fact a quadratic equation, but C (as usual C) is in "quadratic form" go with that one

- 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

@misty1212

- misty1212

the middle term is has \(x^6\) so you can use \(u=x^6\)

- misty1212

ooh what a surprise, C again...

- 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

wow not C

- misty1212

you can use \(u=(x^3+1)\)

- anonymous

What are the solutions of the equation x^4 + 3x^2 + 2 = 0? Use u substitution to solve.

- 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

is there a typo in the question?

- 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

oh they allow complex solutions ok

- anonymous

yes it just took me a minute to type out! :]

- misty1212

\[u+1=00\\
u=-1\\
^2=-1\\
x=\pm i\] is one pair

- misty1212

the other is \(x=\pm\sqrt{2}i\)

- misty1212

looks like they put the \(i\) first , go with B

- anonymous

What are the solutions of the equation x^4 – 5x^2 – 14 = 0? Use factoring to solve.
Same as before.

- 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

then
\[x^2=7\\
x=\pm\sqrt7\] or
\[x^2=-2\\
x=\pm\sqrt{2}i\]

- anonymous

yayy thats what i had :)

- anonymous

thank you!

- misty1212

one to go?

- misty1212

or was that it?

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