## cdelomas Group Title a table for y=2x^2-8 is given solve each equation. a. 2x^2-8=0,b. 2x^2-8<0,c. 2x^2-8>0. one year ago one year ago

1. sauravshakya

where is the table?

2. cdelomas

it is in the question

3. sauravshakya

2x^2-8=0 2x^2=8 x^2=4 x=-2 or 2

4. sauravshakya

got it?

5. cdelomas

is that forthe equation a.

6. sauravshakya

yes

7. cdelomas

8. sauravshakya

Did u understand it?

9. cdelomas

yes for thef irst one I did

10. sauravshakya

2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 Now, IF x^2<(+-a)^2 then -a<x<a

11. sauravshakya

SO, can u complete it?

12. cdelomas

no do not get that one

13. sauravshakya

2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 IS it OKAY up to here?

14. cdelomas

yes

15. cdelomas

would it be 4

16. sauravshakya

IF x^2<(+-a)^2 then -a<x<a

17. sauravshakya

try to use it

18. cdelomas

I do notg et it

19. sauravshakya

x^2<(+-2)^2 -2<x<2

20. sauravshakya

got it?

21. cdelomas

yes but how do i figure it out

22. sauravshakya

HINT: IF x^2<(+-a)^2 then -a<x<a

23. sauravshakya

It is like formula

24. cdelomas

howdo i figurethef o rmula out

25. sauravshakya

U want me to derive it?

26. cdelomas

what does that mean

27. cdelomas

hello what doyou m ean

28. sauravshakya

I am sorry but I guess u need to revise your algebra once

29. cdelomas

soyoucannoth elp meany further

30. sauravshakya

Sorry not so good at explaining things

31. cdelomas

ok Iwil l try someelse

32. sauravshakya

@mathslover can help u?

33. sauravshakya

He is good at explaining

34. mathslover

I am well at explanation :)

35. cdelomas

socanyouhelpme mathsolver

36. mathslover

I will be right back ... wait for 2 minutes please

37. wio

$2x^2-8<0 \\ 2x^2<8 \\ x^2<4 \\ \sqrt{x^2}<\sqrt{4}$Now it is important to remember that $$\sqrt{\quad}$$ mean the "positive square root" (not the negative one) and that $$\sqrt{a^2}$$ is the definition of of the absolute value.$|x| < 2$

38. wio

To solve absolute value equations, we have to split it up into two cases. Case 1: Assume $$x$$ (the thing in the absolute value) is positive. $|x| < 2 \\ x<2$ That was easy. Case 2: Assume $$x$$ is negative: $|x| < 2 \\ -x < 2 \\ x > -2$Remember that if $$x$$ is negative, then $$|x| =-x$$ because the absolute value bars had to flip the sign to make $$x$$ positive. Also remember that when you multiply/divide both sides of an inequality by a negative number, the equality flips. Hence $$<$$ became $$>$$ in this instance.

39. wio

Notation allows us to conveniently combine the expressions $$x < 2$$ and $$x > -2$$ into $$-2 < x < 2$$. Does this help, @cdelomas ?

40. wio

One thing worthy of committing to memory is that: $|x| < a \implies -a < x < a \\ |x| > a \implies x < -a,\quad x > a$

41. cdelomas

so for equation b is it squareroot x2>squareroot 4.