Use the square root property to solve 2d^2 = 3h for d

[kimbrlyar]

Use the square root property to solve 2d[sup:2xdjb7lf]2[/sup:2xdjb7lf] = 3h for d.

I'm not sure where to start?

 

[Deleted member 4993]

Re: Use the square root property to solve for d

2d[sup:1n1kojmf]2[/sup:1n1kojmf] = 3h

I'm not sure where to start?

Can you solve for 'd' - if the equation was:

d[sup:1n1kojmf]2[/sup:1n1kojmf] = h

 

[kimbrlyar]

Re: Use the square root property to solve for d

this is what I did, but I'm not sure, I think I'm missing a step:

2d[sup:1lbjbezl]2[/sup:1lbjbezl]=3h
2d[sup:1lbjbezl]2[/sup:1lbjbezl]/2-3h/2=0
d[sup:1lbjbezl]2[/sup:1lbjbezl]-3h/2=0
d[sup:1lbjbezl]2[/sup:1lbjbezl]-3h/2 * 2/1=0
d[sup:1lbjbezl]2[/sup:1lbjbezl]-6h/2=0
d[sup:1lbjbezl]2[/sup:1lbjbezl]=6h/2
SQRT of d[sup:1lbjbezl]2[/sup:1lbjbezl]=SQRT of 6h/2
d=SQRT of 6h/2

 

[stapel]

2d[sup:12sksqzr]2[/sup:12sksqzr]=3h
2d[sup:12sksqzr]2[/sup:12sksqzr]/2-3h/2=0

You're needing to get "d" by itself, and the instructions specify that you'll be taking the square root of either side, so it would probably be better to leave the "h" term on the other side, away from the "d" term. :idea:

d[sup:12sksqzr]2[/sup:12sksqzr]-3h/2=0
d[sup:12sksqzr]2[/sup:12sksqzr]-3h/2 * 2/1=0

What was your basis for multiplying the one term by 2, but none of the others...? This gives you a new and different equation, with a different solution. :shock:

d[sup:12sksqzr]2[/sup:12sksqzr]=6h/2
SQRT of d[sup:12sksqzr]2[/sup:12sksqzr]=SQRT of 6h/2

Isn't there supposed to be a "plus / minus" on one side of the equation when you take the square root of both sides...?

Note: Check to see if you need to "rationalize" the denominator. Not all instructors require this, but those that do, tend to count off if it's omitted! :wink:

Eliz.

 

[kimbrlyar]

I don't know where to start.

 

[Mrspi]

I don't know where to start.

Start by getting d[sup:tn9qt3mz]2[/sup:tn9qt3mz] by itself on one side of the equals sign, so you'll have an equation that looks like this:

d[sup:tn9qt3mz]2[/sup:tn9qt3mz] = "something"

Then take the square root of both sides....

Please carefully re-read the extensive help you've been given in previous posts.

If you STILL "don't know where to start," then you need more help than we can give you here. You'll need to talk to your teacher.

 

[kimbrlyar]

The answer is d=SQRT of 6h/m.
The work I performed above is the only way I could figure out to get this answer.

I forgot to show it but I did square the "d".

 

[kimbrlyar]

Okay, let me see if this is better.

2d[sup:aszfuw8z]2[/sup:aszfuw8z]/2=3h/2

d[sup:aszfuw8z]2[/sup:aszfuw8z]=3h/2

SQRT d[sup:aszfuw8z]2[/sup:aszfuw8z] = +/- SQRT 3h/2

d = +/- SQRT 3h/2

I'm lost as to how why the right side of the equation is suppose to be +/- SQRT 6h/2 ???

What am I missing? If I rationalize then my denominator would be 4? Again, the answer is d=SQRT 6h/2.

 

[skeeter]

\(\displaystyle d = \pm \sqrt{\frac{3h}{2}} = \pm \frac{\sqrt{3h}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \pm \frac{\sqrt{6h}}{2}\)

 

[stapel]

I'm lost as to how why the right side of the equation is suppose to be +/- SQRT 6h/2 ?

Okay, you need lessons on how to solve by take square roots. Many great lessons are available online:

. . . . .Google results for "solving taking square roots side"

As these lessons will explain and demonstrated, just as the solution to x[sup:bf58u79w]2[/sup:bf58u79w] - 9 = (x - 3)(x + 3) = 0 is not just sqrt[9] = 3, but +/- sqrt[9] = +/- 3, so also the solution to x[sup:bf58u79w]2[/sup:bf58u79w] = m, where "m" is any number, is not just sqrt[m], but is +/- sqrt[m].

If I rationalize then my denominator would be 4? Again, the answer is d=SQRT 6h/2.

I don't know which, if any, of the following you mean...?

. . . . .d = sqrt[6]h/2

. . . . .d = sqrt[6h]/2

. . . . .d = sqrt[6h/2]

But your actual answer was correct: If d[sup:bf58u79w]2[/sup:bf58u79w] = (3h)/2, then d must equal +/- sqrt[(3h)/2].

To learn how to rationalize denominators (your class was supposed to have covered this topic before reaching this stage, too), try here:

. . . . .Google results for "rationalizing denominators radicals"

Once you've studied some lessons, you should understand the following:

Dealing just with the radical expression, you've got sqrt[(3h)/2]; that is, you've got a radical with a fraction inside. Split the radical into a fraction of radicals. You can't simplify the denominator yet, since of course 2 is not a perfect square. But 4 is. So if you multiply by sqrt[2]/sqrt[2], you won't have changed the value at all (since anything divided by itself is just 1, and multiplying by 1 doesn't change anything), and you will be able to simplify the denominator to a whole number!

Eliz.

 

[kimbrlyar]

I appreciate your attention to my dilema.... :?