Solving an Equation w/ Radical

[Giulie]

We just finished up studying quadratic formulas and have now moved onto solving other types of equations. I need some help with some problems. Here are the first few
(this sign ? = square root symbol) :

1. 3?1-X=-2 (the square root is over the entire equation of 1-X)


2. ?8-2x=2x-2 (square root is over entire equation of 8-2x)

Help please!!!

 

[pka]

1. 3?1-X=-2 (the square root is over the entire equation of 1-X)
2. ?8-2x=2x-2 (square root is over entire equation of 8-2x)

So we don't guess are these the problems?
\(\displaystyle \sqrt[3]{{1 - x}} = - 2\;\& \,\sqrt {8 - 2x} = 2x - 2\)

 

[wjm11]

We just finished up studying quadratic formulas and have now moved onto solving other types of equations. I need some help with some problems. Here are the first few
(this sign ? = square root symbol) :

1. 3?1-X=-2 (the square root is over the entire equation of 1-X)


2. ?8-2x=2x-2 (square root is over entire equation of 8-2x)

3?(1-X) = -2
Divide both sides by 3:
?(1-X) = -2/3
Square both sides:
(1-X) = 4/9
Subtract 1 from both sides:
-X = 4/9 – 1= -5/9
Divide (or multiply, same result) both sides by –1:
X = 5/9

ALWAYS check answer by plugging into original equation:
3?(1-(5/9)) = -2
3?(4/9) = -2
3(2/3) = -2
2 = -2 !!!
Our check does NOT work.

Conclusion: this problem has no solution.

Could we have figured out that there was no solution earlier? Yes. When we were at the step ?(1-X) = -2/3, we had a square root on the left side and a negative number on the right side. A square root is always positive, so it was not possible for it to equal a negative number.

You can try the second problem now by using these steps as an example.

 

[pka]

3?(1-X) = -2

WJM11, You should know that there is no sloution to \(\displaystyle 3\sqrt {1 - x} = - 2\).
Because \(\displaystyle \sqrt {1 - x}\) is never negative.

 

[Giulie]

1. 3?1-X=-2 (the square root is over the entire equation of 1-X)
2. ?8-2x=2x-2 (square root is over entire equation of 8-2x)

So we don't guess are these the problems?
\(\displaystyle \sqrt[3]{{1 - x}} = - 2\;\& \,\sqrt {8 - 2x} = 2x - 2\)

yes! those are the problems.

 

[pka]

1. 3?1-X=-2 (the square root is over the entire equation of 1-X)
2. ?8-2x=2x-2 (square root is over entire equation of 8-2x)

So we don't guess are these the problems?
\(\displaystyle \sqrt[3]{{1 - x}} = - 2\;\& \,\sqrt {8 - 2x} = 2x - 2\)

yes! those are the problems.

Then the first it simply cubing both sides: \(\displaystyle 1 - x = - 8\)

The second by squaring: \(\displaystyle 8 - 2x = \left( {2x - 2} \right)^2\).
Be sure to check for extraneous roots.

 

[Giulie]

For the first problem I got x=9


For the second problem I get

4x[supp05rdcx]2[/supp05rdcx]-6x-4=0

after that I am not sure how to continue solving it. I don't believe that factors (but I could be completely wrong) so I intended on creating a quadratic equation now...by plugging the numbers into the quadratic formula....I am not sure if this is a move in the right direction...

 

[pka]

\(\displaystyle \begin{array}{l} 4x^2 - 6x - 4 = 0 \\ 2x^2 - 3x - 2 = 0 \\ (2x + 1)(x - 2) = 0 \\ \end{array}\)

 

[Giulie]

Thank you, thank you, thank you!

so the answers to the second problem are -1/2 or 2.......?

 

[pka]

Thank you, thank you, thank you!
so the answers to the second problem are -1/2 or 2.......?

BE CAREFUL.
Remember I told you to check for extraneous roots.
One of those two will not work.