Solving equations

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rachelmaddie

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Hi. Can some please check my work?
Solve
6 + 5*sqrt(249 - 2x) = 7.

sqrt(249 - 2x) = 1/5 =>
249 - 2x = 1/25 =>
2x = 248.96 and x = 124.48.
 
BIG HINT

[MATH]6 + \sqrt[5]{249 - 2x} = 7 \implies[/MATH]

[MATH]\sqrt[5]{249 - 2x} = 7 - 6 = 1.[/MATH]
 
JeffM said:
BIG HINT

[MATH]6 + \sqrt[5]{249 - 2x} = 7 \implies[/MATH]

[MATH]\sqrt[5]{249 - 2x} = 7 - 6 = 1.[/MATH]
Click to expand...
Hows this?
Solve
6 + 5*√249 - 2x) = 7 - 6
(5*√249 - 2x)^5 =(1)^5
249 - 2x = 1 —> -
-2x/-2 = -248/2 = 124
 
rachelmaddie said:
Hows this?
Solve
6 + 5*√249 - 2x) = 7 - 6
(5*√249 - 2x)^5 =(1)^5
249 - 2x = 1 —> -
-2x/-2 = -248/2 = 124
Click to expand...
That is fine except that you should check your work.

[MATH]6 + \sqrt[5]{249 - 2 * 124}=\\ 6 + \sqrt[5]{249 - 248} = \\ 6 + \sqrt[5]{1} = 6 + 1 = 7 \checkmark.[/MATH]
 
JeffM said:
That is fine except that you should check your work.

[MATH]6 + \sqrt[5]{249 - 2 * 124}=\\ 6 + \sqrt[5]{249 - 248} = \\ 6 + \sqrt[5]{1} = 6 + 1 = 7 \checkmark.[/MATH]
Click to expand...
Solve
6 + 5*√249 - 2x) = 7

Subtract 6 on each side. This gives you:

5*√249 - 2x = 1

Raise each side to the power of 5.

(5*√249 - 2x)^5 = (1)^5

Cancel the 5 out on the left side and the root sign.

249 - 2x = 1

Subtract 249 on each side of the equation. This gives you
-2x = -248

Lastly, divide by -2 on both sides. This gives you
x = 124

Therefore, your answer is x = 124.
 
TO rachelmaddie: NO, NO, NONO.
first please learn some basic LaTeX. \sqrt[5]{249-2x}\ gives \(\sqrt[5]{249-2x}\)
Can you solve \(249-2x=1~?\)
 
First 6 + 1 = 7
So [math]\sqrt[5]{249-2x} = 1[/math]Then 249-2x = 1
So 2x=248
Or x = 124.
 
rachelmaddie said:
Solve
6 + 5*√249 - 2x) = 7
Click to expand...
It is not 5 times the square root of (249 - 2x), which is what you wrote above. It is the fifth root of (249 - 2x). Your notation is wrong. Because you do not know LaTeX, you need to write something like

6 + fifth root of (249 - 2x) = 7 OR, if you know about fractional exponents 6 + (249 - 2x)^(1/5)

Subtract 6 on each side.
Click to expand...

Correct

This gives you:

5*√249 - 2x = 1
Click to expand...

Same misuse of notation as above

So fifth root of (249 - 2x) = 1 OR (249 - 2x)^(1/5) =1

Raise each side to the power of 5.

(5*√249 - 2x)^5 = (1)^5
Click to expand...

You have the right idea but the wrong notation.

Raise the fifth root of (249 - 2x) to the fifth power = 1^ 5or {(249 - 2x)^(1/5)}^5 = 1^5

Cancel the 5 out on the left side and the root sign.

249 - 2x = 1
Click to expand...

You have the right idea, but the wrong notation. You cannot cancel powers and coefficients.

Subtract 249 on each side of the equation. This gives you
-2x = -248

Lastly, divide by -2 on both sides. This gives you
x = 124

Therefore, your answer is x = 124.
Click to expand...
I believe you had the right idea, but your notation will suggest to people that you got the right answer by accident.

I suspect that you are about 14 or 15. If that is right, I cannot support pka's suggestion that you learn LaTeX, which is a very fussy language, when you are having trouble understanding the algebra that you are trying hard to learn. One thing at a time. But you do need to be careful with your notation. Otherwise no one will understand what you are saying.
 
rachelmaddie said:
5*sqrt(249 - 2x)
Click to expand...
Hi Rachel. One way to text radicals is to type the index inside brackets:

root[5](249 - 2x)

Whichever notation you choose, it's good form to define it at the beginning:

"Note: root[n](number) designates the nth root of number", or something like that.

?
 
JeffM said:
It is not 5 times the square root of (249 - 2x), which is what you wrote above. It is the fifth root of (249 - 2x). Your notation is wrong. Because you do not know LaTeX, you need to write something like

6 + fifth root of (249 - 2x) = 7 OR, if you know about fractional exponents 6 + (249 - 2x)^(1/5)



Correct



Same misuse of notation as above

So fifth root of (249 - 2x) = 1 OR (249 - 2x)^(1/5) =1



You have the right idea but the wrong notation.

Raise the fifth root of (249 - 2x) to the fifth power = 1^ 5or {(249 - 2x)^(1/5)}^5 = 1^5



You have the right idea, but the wrong notation. You cannot cancel powers and coefficients.


I believe you had the right idea, but your notation will suggest to people that you got the right answer by accident.

I suspect that you are about 14 or 15. If that is right, I cannot support pka's suggestion that you learn LaTeX, which is a very fussy language, when you are having trouble understanding the algebra that you are trying hard to learn. One thing at a time. But you do need to be careful with your notation. Otherwise no one will understand what you are saying.
Click to expand...
No, I’m not. I’m 19
 
JeffM said:
It is not 5 times the square root of (249 - 2x), which is what you wrote above. It is the fifth root of (249 - 2x). Your notation is wrong. Because you do not know LaTeX, you need to write something like

6 + fifth root of (249 - 2x) = 7 OR, if you know about fractional exponents 6 + (249 - 2x)^(1/5)



Correct



Same misuse of notation as above

So fifth root of (249 - 2x) = 1 OR (249 - 2x)^(1/5) =1



You have the right idea but the wrong notation.

Raise the fifth root of (249 - 2x) to the fifth power = 1^ 5or {(249 - 2x)^(1/5)}^5 = 1^5



You have the right idea, but the wrong notation. You cannot cancel powers and coefficients.


I believe you had the right idea, but your notation will suggest to people that you got the right answer by accident.

I suspect that you are about 14 or 15. If that is right, I cannot support pka's suggestion that you learn LaTeX, which is a very fussy language, when you are having trouble understanding the algebra that you are trying hard to learn. One thing at a time. But you do need to be careful with your notation. Otherwise no one will understand what you are saying.
Click to expand...
Is this better?
Solve
6 + fifth root of (249 - 2x) = 7

Subtract 6 on each side. This gives you:

fifth root of (249 - 2x) = 1

Raise each side to the power of 5.

Raise the fifth root of (249 - 2x) to the fifth power = 1^ 5

249 - 2x = 1

Subtract 249 on each side of the equation. This gives you
-2x = -248

Lastly, divide by -2 on both sides. This gives you
x = 124

Therefore, your answer is x = 124.
 
That exactly is the logic needed to solve this equation. Very good.

When working with an unknown inside a root, a technique that frequently works is to get the root all by itself on one side of the equation (isolate the root), and then raise BOTH sides of the equation by the power that eliminates the root. Then you have a simpler equation without any root.

You follow that general process?
 
JeffM said:
That exactly is the logic needed to solve this equation. Very good.

When working with an unknown inside a root, a technique that frequently works is to get the root all by itself on one side of the equation (isolate the root), and then raise BOTH sides of the equation by the power that eliminates the root. Then you have a simpler equation without any root.

You follow that general process?
Click to expand...
Which equation?
 
In one sense, the MECHANICS of algebra are about moving step by step from a complex equation to a very simple equation.

[MATH]6 + \sqrt[5]{249 - 2x} = 7.[/MATH] Original complex equation

[MATH]\sqrt[5]{249 - 2x} = 1.[/MATH] Simpler. Root is isolated by itself.

[MATH](\sqrt[5]{249 - 2x})^5 = 1^5.[/MATH] Raise both sides of previous equation to 5th power

[MATH]249 - 2x = 1.[/MATH] Simpler now for sure because the root was eliminated.

[MATH]249 - 1 = 2x.[/MATH] Unknown is isolated.

[MATH]2x = 248.[/MATH] Very simple equation.

[MATH]x = 124.[/MATH]
Solving an algebraic equation involves taking steps to arrive at the super-simple equation x = some numeral.
 
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