## anonymous one year ago 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)

1. misty1212

lol HI again!

2. anonymous

hey girl!

3. misty1212

$(x + 2)^2 + 5(x + 2) – 6 = 0$ put $$\color{red}u=x+2$$ get $\huge\color{red}u^2+5\color{red}u-6=0$

4. anonymous

or just go with C since it is always C

5. misty1212

lol ikr!

6. anonymous

When in doubt pick C lol.

7. 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

8. anonymous

@misty1212

9. misty1212

quadratic means hmm this is a trick

10. 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 $10u^2+7u+1=0$

11. misty1212

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

12. 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

13. anonymous

@misty1212

14. misty1212

the middle term is has $$x^6$$ so you can use $$u=x^6$$

15. misty1212

ooh what a surprise, C again...

16. 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

17. misty1212

wow not C

18. misty1212

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

19. anonymous

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

20. 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

21. misty1212

is there a typo in the question?

22. 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

23. misty1212

oh they allow complex solutions ok

24. anonymous

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

25. misty1212

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

26. misty1212

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

27. misty1212

looks like they put the $$i$$ first , go with B

28. anonymous

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

29. misty1212

yeah same as before $u=x^2$ get $u^2-5u-14=0$ or $(u-7)(u+2)=0$ so $u=7$or $u=-2$

30. misty1212

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

31. anonymous

yayy thats what i had :)

32. anonymous

thank you!

33. misty1212

one to go?

34. misty1212

or was that it?