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Use Square Root Property to Solve 3x^2 - 1 = 47
- Thread starter tmcintyre
- Start date
D
Deleted member 4993
Guest
Re: Square Root
tmcintyre said:That is my problem. It says use the square root property to solve the equation. If necessary, round to two decimal places. I came up with 7. ? That is incorrect.
If you put x = 7 in your given equation you'll see it is not satisfied - as follows
3 * x[sup:250falkk]2[/sup:250falkk] - 1 = 47
3 * 7[sup:250falkk]2[/sup:250falkk] - 1 = 47
147 - 1 = 47
146 = 47 ? obviously not satisfied
How did you get x = 7 ??
Please show us your work - so that we know where to begin to help
Re: Square Root
Can you solve this: 2x + 1 = 33 ?
You're obviously "lost", unaware of the basics.tmcintyre said:To be honest with you, I dont know how I came up with 7. I did the problem again and came up with:
3x^2-1=47
3x-1
x^2=1/3
x=1/3
That is how my book shows me to do this problem, but I dont know if my answer is right.
Can you solve this: 2x + 1 = 33 ?
D
Deleted member 4993
Guest
Re: Square Root
I cannot follow what you have tried to do. For example:
How did line 2 follow from line 1?
where did the '=' sign go in line 2?
where did the "^2" go in line 2?
How did 3 follow from line 2?
tmcintyre said:To be honest with you, I dont know how I came up with 7. I did the problem again and came up with:
3x^2-1=47
3x-1
x^2=1/3
x=1/3
That is how my book shows me to do this problem, but I dont know if my answer is right.
I showed you above - how to check, if your answer is right. Put your answer back into the given equation, and see if the equation is satisfied.
I cannot follow what you have tried to do. For example:
How did line 2 follow from line 1?
where did the '=' sign go in line 2?
where did the "^2" go in line 2?
How did 3 follow from line 2?
Re: Square Root
Hi tmcintyre,
I would hope that you have some examples from class to go by. The first thing to do would be to add 1 to both sides.
\(\displaystyle 3x^2-1=47\)
\(\displaystyle 3x^2-1+1=47+1\)
\(\displaystyle 3x^2=48\)
Next, in order to isolate the variable, we must divide both sides by 3.
\(\displaystyle \frac{3x^2}{3}=\frac{48}{3}\)
\(\displaystyle x^2=16\)
Now, you see that both sides of the equation are perfect squares. Take the square root of each side and you're done.
\(\displaystyle \sqrt{x^2}=\sqrt{16}\)
\(\displaystyle x= \pm 4\)
To check, simply substitute your answers back into you original equation to see if they work.
For further referece, look at some worked exampes here: http://www.cliffsnotes.com/WileyCDA/Cli ... 38933.html
tmcintyre said:To be honest with you, I dont know how I came up with 7. I did the problem again and came up with:
3x^2-1=47
3x-1
x^2=1/3
x=1/3
That is how my book shows me to do this problem, but I dont know if my answer is right.
Hi tmcintyre,
I would hope that you have some examples from class to go by. The first thing to do would be to add 1 to both sides.
\(\displaystyle 3x^2-1=47\)
\(\displaystyle 3x^2-1+1=47+1\)
\(\displaystyle 3x^2=48\)
Next, in order to isolate the variable, we must divide both sides by 3.
\(\displaystyle \frac{3x^2}{3}=\frac{48}{3}\)
\(\displaystyle x^2=16\)
Now, you see that both sides of the equation are perfect squares. Take the square root of each side and you're done.
\(\displaystyle \sqrt{x^2}=\sqrt{16}\)
\(\displaystyle x= \pm 4\)
To check, simply substitute your answers back into you original equation to see if they work.
For further referece, look at some worked exampes here: http://www.cliffsnotes.com/WileyCDA/Cli ... 38933.html