Solving a Square Root Equation with Fractional Coefficients

In summary, this conversation is discussing how to solve an equation involving the square roots of x. The first equation gets rid of the square root signs and then the second equation cross multiplies the left hand side. The final answer is 5= sqrt (49- 9x - 42 sqrt x)
  • #1
Quinn Morris
14
0

Homework Statement


[tex]\frac{5}{\sqrt{7+3\sqrt{x}}}[/tex] = [tex]\sqrt{7 -3\sqrt{x}}[/tex]


Homework Equations



none

The Attempt at a Solution


does this equal 5 = 7 - 3[tex]\sqrt{x}[/tex]
 
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  • #2
i cross multiplied, but I'm a little iffy on my FOIL technique when square roots are throw into the mix
 
  • #3
First square both sides and then cross multiply. That's what i'd do.
 
  • #4
5 = sqrt (49 - 9x - 42 sqrt x)
25 = 49 - 9x - 42 sqrt x
24 = 9x + 42 sqrt x
The rest I leave it to you. You just have to square another time.
 
  • #5
i don't understand how you got that could you explain it please?
 
  • #6
lkh1986 said:
5 = sqrt (49 - 9x - 42 sqrt x)
25 = 49 - 9x - 42 sqrt x
24 = 9x + 42 sqrt x
The rest I leave it to you. You just have to square another time.

This term shouldn't be there. Check your signs.
 
Last edited:
  • #7
okay so square both sides to get rid of the first square root sign then square again to get rid of the one on the "x" ?
 
  • #8
Quinn Morris said:
i cross multiplied, but I'm a little iffy on my FOIL technique when square roots are throw into the mix

Square both sides (this gets rid of the "first set" of square roots). Then cross multiply. Expand the right hand side. Note your laws of exponents: so [tex] x^\frac{1}{2} * x^\frac{1}{2} = x ^ { \frac{1}{2} + \frac{1}{2}} = x^1 = x[/tex]
 
  • #9
okay I'm definitely lost now.

i have the stated question

i sqaure both sides, i get

[tex]\frac{25}{7 + 2((\sqrt{7})(\sqrt{3\sqrt{x}})) + 3\sqrt{x}}[/tex] = 7 + 2(([tex]\sqrt{7}[/tex])([tex]\sqrt{-3\sqrt{x}}[/tex])) + 3[tex]\sqrt{x}[/tex]
 
  • #10
how can i cross multiply that without getting utterly lost?
 
  • #11
By square both sides, I mean


[tex](\frac{5}{\sqrt{7+3\sqrt{x}}})^2[/tex] = [tex](\sqrt{7 -3\sqrt{x}} )^2[/tex]

This means squaring every term

[tex]\frac{5^2}{(\sqrt{7+3\sqrt{x}})^2}[/tex] = [tex](\sqrt{7 -3\sqrt{x}} )^2[/tex]

Can you tell me what this is equivalent to: [tex](\sqrt{x})^2[/tex]? If you know this, then you will know how to make the above much easier to work with. Figure this out before you expand anything. See my laws of exponents remark.

By the way the curved brackets in the second equation go around the whole square root term. Hopefully it doesn't look too confusing to you.
 
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  • #12
sqrt x squared is just X. but i get confused with the middle terms. I am going to try it later on when i have the time to, i'll get back to you
 
  • #13
Yes, that's right. So if you have

[tex](\sqrt{7 -3\sqrt{x}} )^2[/tex]

and apply that idea, what does this term become?
 
  • #14
The parentheses are peculiarly placed!
Multiplying
[tex]\frac{5}{\sqrt{7+3\sqrt{x}}}= \sqrt{7 -3\sqrt{x}}[/tex]
on both sides by [itex]\sqrt{7+ 3\sqrt{x}}[/itex] gives [itex]5= \sqrt{(7- 3\sqrt{x})(7+ 3\sqrt{x})}= \sqrt{49- 9x}[/itex], because the product inside the squareroot is "sum times difference". Now squaring gives 25= 49- 9x. That should be easy to solve. Don't forget to check your answer- multiplying both sides by something involving x or squaring both sides can introduce "extraneous" solutions.
 
  • #15
c=8 i think
 
  • #16
x = 8/3 I think.
 
  • #17
Quinn Morris said:
c=8 i think
Since there isno "c" in the problem, I guess you missed the "x" key!
Unfortunately, with x= 8, the [itex]7- 3\sqrt{8}[/itex] is clearly negative and so cannot satisfy the equation.

epenguin said:
x = 8/3 I think.
Check it!
[tex]\frac{5}{\sqrt{7+ 3\sqrt{8/3)}}}= 1.4494897427831780981972840747059[/tex]
while
[tex]\sqrt{7- 3\sqrt{(8/3)}}= 1.4494897427831780981972840747059[/tex]
Looks good to me!

As I said before,
HallsofIvy said:
Multiplying
[tex]\frac{5}{\sqrt{7+3\sqrt{x}}}= \sqrt{7 -3\sqrt{x}}[/tex]
on both sides by [itex]\sqrt{7+ 3\sqrt{x}}[/itex] gives [itex]5= \sqrt{(7- 3\sqrt{x})(7+ 3\sqrt{x})}= \sqrt{49- 9x}[/itex], because the product inside the squareroot is "sum times difference". Now squaring gives 25= 49- 9x. That should be easy to solve. Don't forget to check your answer- multiplying both sides by something involving x or squaring both sides can introduce "extraneous" solutions.
If 25= 49- 9x, the 9x= 49-25= 24 so x= 24/9= 8/3 as epenguin said. (But you should still check numerically.)
 
  • #18
lol i just rechecked my work and I'm an idiot. i had 24/9 but i divded both by 3 but the i just decided not to include the /3 part. lol @ me thanks for the help tho
 

1. What is a square root?

A square root is a number that, when multiplied by itself, gives the original number. For example, the square root of 25 is 5 because 5 multiplied by itself equals 25.

2. How do you calculate the square root of a number?

To calculate the square root of a number, you can either use a calculator or use long division. The long division method involves finding the largest square number that is less than or equal to the given number and then finding the difference between the given number and the square number. This process is repeated until the desired level of accuracy is achieved.

3. What is the difference between a perfect square and a non-perfect square?

A perfect square is a number whose square root is a whole number. For example, 25 is a perfect square because its square root is 5. A non-perfect square is a number whose square root is a decimal or irrational number. For example, the square root of 2 is a non-perfect square because it is an irrational number (1.414213...).

4. Can you have a negative square root?

Yes, you can have a negative square root. In fact, every positive number has two square roots - one positive and one negative. For example, the square root of 25 is both 5 and -5.

5. How are square roots used in real life?

Square roots are used in many real-life situations, such as calculating the side lengths of a square or finding the dimensions of a square room. They are also used in advanced mathematical concepts, such as in the Pythagorean theorem and in calculus.

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