Square roots/radicals equations

A

Audentes

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I did the Algebra One unit on square roots and radicals in the fall, and I was doing good. Now, I got a review for a test on everything in the year so far, and the questions are much more complicated than the ones I were given a few months ago.

I need help with these two:

[MATH](x/2)=sqrt(3x)[/MATH]
[MATH]x=sqrt(2x+3)[/MATH]
I keep feeling like the math makes me go in circles with no end or proper substitute for the variable...
 
You stated two equations but failed to say what you wanted to do to them.

1)You may want to multiply both sides by 11.
2) You want to add 2 to both sides.
3) You want to compute the cube of both sides.
4) You may want to solve for x.
....

Please finish you post by asking a question.
 
Flo Rida said:
[MATH](x/2)=sqrt(3x)[/MATH]
[MATH]x=sqrt(2x+3)[/MATH]
I keep feeling like the math makes me go in circles with no end or proper substitute for the variable...
Click to expand...
Please show us the circles you are in, so we can correct your compass.

As it is, I can see several things you could be doing wrong (and several things you could be doing right!).
 
Jomo said:
You stated two equations but failed to say what you wanted to do to them.

1)You may want to multiply both sides by 11.
2) You want to add 2 to both sides.
3) You want to compute the cube of both sides.
4) You may want to solve for x.
....

Please finish you post by asking a question.
Click to expand...
Dr.Peterson said:
Please show us the circles you are in, so we can correct your compass.

As it is, I can see several things you could be doing wrong (and several things you could be doing right!).
Click to expand...
For the first one, I would like to multiply both sides by 2 to get x alone.
gives me x= sqrt(3x)*2


for the second one, I would like to square each side.
that leaves me with x^2 = 2x+3
then when I divide that by x, I get x=2+(3/x).

for both of these, after the last step I show here, I can’t seem to get the answer for x
 
The question you should have asked was how do I solve for x.

x^2 = 2x+3 is a quadratic equation. You should not divide by x. You should use one of the few methods which you learned to solve quadratics. The first step is to get one side of the equation to be 0.

x= sqrt(3x)*2. Now square both sides. Then you'll have a quadratic equation ....
 
Last edited:
Jomo said:
x= sqrt(3x)*2. Now square both sides. Then you'll have a quadratic equation ....
Click to expand...
this gives me x^2=(3x)*4, which can become x^2=12x. Divide both sides by x and I get x=12. That is the answer?
 
Jomo said:
x^2 = 2x+3 is a quadratic equation. You should not divide by x. You should use one of the few methods which you learned to solve quadratics. The first step is to get one side of the equation to be 0.
Click to expand...
My teacher barely did quadratics so far, but I’ll try this.

UPDATE: I subtracted the x on the left side and now have 0=sqrt(2x+3)-x
 
Flo Rida said:
this gives me x^2=(3x)*4, which can become x^2=12x. Divide both sides by x and I get x=12. That is the answer?
Click to expand...
Quadratics could have up to 2 answers. Can you find another solution? Can you think of a number (other than 12) that when you multiply it by itself you get the same answer if you multiplied this number by 12?
 
Jomo said:
Quadratics could have up to 2 answers. Can you find another solution? Can you think of a number (other than 12) that when you multiply it by itself you get the same answer if you multiplied this number by 12?
Click to expand...
Zero? Because I think that if I subsitute zero both sides should be 0.
 
Flo Rida said:
Zero? Because I think that if I subsitute zero both sides should be 0.
Click to expand...
Why are you not sure? Is 0*0 = 0*12? If it is then x=0 is a solution otherwise it is not.

Now why did you not get that solution on your own? The answer is because you divided both sides bye. You should have factored out an x instead of dividing by it!
 
Jomo said:
Why are you not sure? Is 0*0 = 0*12? If it is then x=0 is a solution otherwise it is not.
Click to expand...
Yes, so then 0 is a solution.
Jomo said:
Now why did you not get that solution on your own? The answer is because you divided both sides bye. You should have factored out an x instead of dividing by it!
Click to expand...
Oh okay, that does make sense
 
Flo Rida said:
My teacher barely did quadratics so far, but I’ll try this.

UPDATE: I subtracted the x on the left side and now have 0=sqrt(2x+3)-x
Click to expand...
No, what you did before, squaring, was correct. Don't back up.

In order to solve these, especially the second, you need to have studied quadratic equations. If you didn't, then they shouldn't be assigned to you. (That is, perhaps you don't have the prerequisites for what you are learning now.)

Does it sound at all familiar that you can solve these by getting a zero on one side and factoring the other? If not, search for "solving quadratic equations by factoring".
 
Dr.Peterson said:
No, what you did before, squaring, was correct. Don't back up.

In order to solve these, especially the second, you need to have studied quadratic equations. If you didn't, then they shouldn't be assigned to you. (That is, perhaps you don't have the prerequisites for what you are learning now.)

Does it sound at all familiar that you can solve these by getting a zero on one side and factoring the other? If not, search for "solving quadratic equations by factoring".
Click to expand...
so I just asked my teacher about this and he said that he will waive these questions and exclude them. he said that while we did cover it in relevance to exponents, etc., didn’t go in depth and actual studying of quadratics is the first unit after this test.
 
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