Question

please help

Answer 1

\[2x-7\sqrt{x}-30=0\]

Answer 2

Just solve for x?

Answer 3

I feel like it would be possibly easier if you square everything

Answer 4

ya i need to solve for x, but i think i slept through this lesson and missed the notes so im not sure what to do

Answer 5

Square both sides so that
\[4x^2-49x-900=0\]
and simplify from there on out

Answer 6

Wait, that;s wrong :S nevermind

Answer 7

Please write down, at least in your mind, \(x \ge 0\). THEN square things as you will.

However, you could just recognize the quadratic form and try to factor it (quadratic formula would work, too)

\(2x - 7\sqrt{x} - 30 = (2\sqrt{x}+5)(\sqrt{x}-6) = 0\)

Answer 8

i really have no idea what your talking about

Answer 9

How would you approach this problem:

\(2w^{2} − 7w − 30 = 0 \) Solve for \(w\).

Answer 10

w^2-7w-60=0

Answer 11

?? That makes no sense. You can't just magically change values.

If you wanted to divide by 2, you should have gotten \(w^{2} - (7/2)w - 15 = 0\). This is not very helpful.

Let's back up another step and ask again, how would you approach this problem?

\(s^2 - s - 6 = 0\)

Answer 12

i would find the factors of -6 and -1

Answer 13

You would factor the left-hand expression?

(s-3)(s+2) = 0 and produce s = 3 or s = -2

Does that make sense?

Answer 14

yes

Answer 15

Excellent. Now back to the other question.

How would you approach this problem statement?

Solve for \(w\) \(2w^{2} - 7w - 30 = 0\)

Please say, "Look at the factors of 30 and the factors of 2 and try to factor the left-hand expression."

Answer 16

w= 3/2 and w=5?

Answer 17

\(2w^{2} - 7w - 30 = (2w+5)(w-6) = 0\) leads to w = -5/2 or w = 6

You just have to be a little more careful. Do you see it, so far?

Answer 18

i guess so -_-

Answer 19

You just have to find the right factorization.

Working backwards, here's yours...

w= 3/2 or w=5

\((2w - 3)(w-5) = 2w^{2} - 13w + 15\) That's not quite where we started, is it?

Answer 20

i really just need to know who to deal with the \[7\sqrt{x}\]

Answer 21

i can take it from thier

Answer 22

I'm trying to tell you that and we're almost there. The important part is this:
Something-squared + Somehting + Constant = 0 calls for factoring. It doesn't matter what the "somehting" is. In your case, "somehting" is \(\sqrt{x}\). Recognize the form and:

1) Factor it \((2\sqrt{7} + 5)(\sqrt{7}-6)\), or

2) Use the Quadratic formula: \(\sqrt{x} = \dfrac{7 \pm \sqrt{(-7)^{2} - 4(2)(-30)}}{2(2)}\)

Answer 23

Sorry, that #1 should have been \((2\sqrt{x}+5)(\sqrt{x}-6)\). I got a little jumpy with the 7s.

Answer 24

so \[\sqrt{x}=\frac{ 7\pm17 }{ 4 }?\]

Answer 25

Correct. Points for algebra. Now, what about x?

Answer 26

Reminder, since that square root mess worked out nicely to 17 we could have factored the original expression!

Answer 27

factored it how?

Answer 28

That's what #1 is.

Answer 29

It is the same as the quadratic formula. Unique answers don't care how you find them.

Answer 30

oh ok
thanks

Answer 31

Okay. Don't forget to discard the negative value. It's no good!