Soving a Two Radical Equation for Roots

[Jason76]

How do I solve this for roots?

\(\displaystyle \sqrt{x+4} + \sqrt{1-x} = 3 \)

Solve for one radical and eliminate.

\(\displaystyle \sqrt{x + 4} = 3 - \sqrt{1 - x}\)

Square both sides to get rid of radicals.

\(\displaystyle \sqrt{x + 4}^2 = (3 - \sqrt{1 -x})^2\)

Squaring both sides (to get rid of radicals) should lead to this (by my logic):

\(\displaystyle x + 4 = 9 - 1 - x\)

But this is the correct answer (according to the book):

\(\displaystyle x + 4 = 9 - 6 \sqrt{1-x} + (1 - x) \)

That's main part I have trouble with. The rest is easy to understand.

Of course, this only half of the problem. I haven't solved for roots yet.

 

[daon2]

There may be several steps needed. In general what you want to do is:

1) solve for a radical
2) square
3) rinse & repeat until no more radicals remain

You will have a quadratic function in the case of two radicals when all said and done. This will produce two potential solutions. You select only those in the domain of your function on the left hand side.

Note that in the case of INTEGER solutions, this particular question is quickly solved. The domain of your function (on the LHS) is \(\displaystyle [-4,1]\). Testing -4, -3, -2, -1, 0, 1 will give you the integer solutions. Since there are two, you're done.

 

[Jason76]

NO! Hint: (a - b)^2 = a^2 - 2ab + b^2

Right, but when you square

\(\displaystyle \sqrt{x + 4} \) and \(\displaystyle 3 - \sqrt{1 - x}\)

\(\displaystyle (\sqrt{x + 4})^2 \) and \(\displaystyle (3 - \sqrt{1 - x})^2\)

then that should get rid of the radical and do no more.

Yields:

\(\displaystyle x + 4\) and \(\displaystyle 3 - 1 - x \)

If you look above, you see no squares available for the "distributive property".

 

[mmm4444bot]

Right, but when you square

\(\displaystyle 3 - \sqrt{1 - x}\)

then that should get rid of the radical and do no more.

That's not correct logic, Jason.

You are not seeing the square of a binomial on the right-hand side, but that is what you have.

Denis showed you the pattern for squaring a binomial.

Look at the equation that Denis posted.

Now, on each side of that equation, substitute a = 3 and b = sqrt(1 - x).

But this is the correct answer (according to the book):

\(\displaystyle x + 4 = 9 - 6 \sqrt{1-x} + (1 - x) \)

Odd answer; I'm wondering what the question was.

Anyway, after you substitute into the binomial-square pattern (posted by Denis) and do the arithmetic, you'll see that your equation simplifies to match the book's.

By the way, have you ever seen the acronym FOIL?

 

[Jason76]

That's not correct logic, Jason.

You are not seeing the square of a binomial on the right-hand side, but that is what you have.

Denis showed you the pattern for squaring a binomial.

Look at the equation that Denis posted.

Now, on each side of that equation, substitute a = 3 and b = sqrt(1 - x).





Odd answer; I'm wondering what the question was.

Anyway, after you substitute into the binomial-square pattern (posted by Denis) and do the arithmetic, you'll see that your equation simplifies to match the book's.

By the way, have you ever seen the acronym FOIL?

Ok in regards to,

\(\displaystyle (3 - \sqrt{1-x})\)

The whole above expression is a binomial. However, the left side expression:

\(\displaystyle \sqrt{x + 4}\)

is NOT a binomial, so it will be simply squared (with the purpose of eliminating the radical sign). Is that right? Ok, so now what's the next step with FOIL? How is the right side squared?

 

[mmm4444bot]

We are not discussing the left-hand side. Try to focus.

a^2 - 2*a*b + b^2

You need to replace symbol a above with the number 3

You need to replace symbol b above with the radical \(\displaystyle \sqrt{1-x}\)

If you do not know how to do this type of substitution, or how to simplify after replacing the symbols, then I think that you need to chat up your instructor. Something has gone wrong with your prerequisites.

Really, FOIL needs to be taught before assigning this type of exercise. That fact that you are unfamiliar with FOIL explains to me why you are unable to see the square of a binomial.

Anyway, do the substitution and simplify (FOIL is not required for that) or talk to your teacher.

 

[Jason76]

We are not discussing the left-hand side. Try to focus.

a^2 - 2*a*b + b^2

You need to replace symbol a above with the number 3

You need to replace symbol b above with the radical \(\displaystyle \sqrt{1-x}\)

If you do not know how to do this type of substitution, or how to simplify after replacing the symbols, then I think that you need to chat up your instructor. Something has gone wrong with your prerequisites.

Really, FOIL needs to be taught before assigning this type of exercise. That fact that you are unfamiliar with FOIL explains to me why you are unable to see the square of a binomial.

Anyway, do the substitution and simplify (FOIL is not required for that) or talk to your teacher.

Ok, now I see what's going on, sort of.

\(\displaystyle (a - b) ^2 = (a^2 - ab + b^2)\)

\(\displaystyle a = 3\)

\(\displaystyle b = \sqrt{1 - x}\)

so regarding the right side:

On the far sides, a is squared and so is b (Therefore, the radical sign is removed). In the middle, the radical term stays the same (because it isn't squared) and a 6 comes up.

However, Where did the 6 come from? Shouldn't the equation be

\(\displaystyle 3 - 3(\sqrt{1 - x}) + (1 - x)\) ?

Also, to correct my former post, binomials exist on both sides of the equation, but only one of them is a "square of a binomial".

 

[mmm4444bot]

Also, to correct my former post, binomials exist on both sides of the equation, but only one of them is a "square of a binomial".

You don't need to explain this to us.

And, the six comes from multiplying 2*3.

You're just not paying attention.

Goodbye.

 

[Jason76]

I think I got it now:

Recognize right hand side of equation as a "Square of a Binomial".

\(\displaystyle (3 - \sqrt{1-x})^2\)

\(\displaystyle (3 - \sqrt{1-x}) (3 - \sqrt{1-x}) \)

FOIL

\(\displaystyle (3)(3) + (3) ( - \sqrt{1-x}) + (3) ( - \sqrt{1-x}) + ( - \sqrt{1-x}) ( - \sqrt{1-x}) \)

\(\displaystyle 9 + (3) ( - \sqrt{1-x}) + (3) ( - \sqrt{1-x}) + ( 1-x ) \)

Next, add the two middle terms. In this case, it's adding negatives, so it's -b - b. The same as -b (+) -b

\(\displaystyle 9 - 3 \sqrt{1-x} - 3 \sqrt{1-x} + ( 1-x ) \)

\(\displaystyle 9 - 6 \sqrt{1-x} + ( 1-x)\)

Full equation:

\(\displaystyle x + 4 = 9 - 6 \sqrt{1-x} + ( 1-x)\)

Next, more algebra until the roots are found.

 

[lookagain]

There may be several steps needed. In general what you want to do is:

1) solve for a radical
2) square
3) rinse & repeat until no more radicals remain

You may ** have a quadratic function in the case of two radicals when all said and done. This will produce two potential solutions. You select only those in the domain of your function on the left hand side.

**For example, \(\displaystyle \sqrt{x - 1} + 2 = \sqrt{x + 3}\) will not end up being

a quadratic function equal to zero.

--------------------------------------------------------



Jason76,

it may be recommended, but it is not required that you subtract one of the radical terms

from each side at the beginning. If you were to FOIL the left-hand side at the beginning,

there would be just this one radical expression, namely \(\displaystyle 2\sqrt{x+4}\sqrt{1 - x}.\)

And after it is isolated, then the squaring on both sides of the equation
(the second time in the problem), will eliminate all of the radical terms
in the equation.

 

[Jason76]

**For example, \(\displaystyle \sqrt{x - 1} + 2 = \sqrt{x + 3}\) will not end up being

a quadratic function equal to zero.

--------------------------------------------------------



Jason76,

it may be recommended, but it is not required that you subtract one of the radical terms

from each side at the beginning. If you were to FOIL the left-hand side at the beginning,

there would be just this one radical expression, namely \(\displaystyle 2\sqrt{x+4}\sqrt{1 - x}.\)

And after it is isolated, then the squaring on both sides of the equation
(the second time in the problem), will eliminate all of the radical terms
in the equation.

Right, but what I did in my last post was correct. However, more steps are needed to get to the roots. However, those are easy, since the "FOIL Thing" was my main stumbling block.

-----

Ok. Here is the 2nd part of the equation:

\(\displaystyle x + 4 = 9 - 6 \sqrt{1-x} + ( 1-x)\)

Isolate the radical again by moving it to the left side.

\(\displaystyle (6 \sqrt{1-x})^2 = (6 - 2x)^2\)

\(\displaystyle 36(1-x) = 66 - 24x + 4x^2\)

Turn this into a quadratic equation by creating a 0 on one side.

\(\displaystyle 4x^2 + 12x = 0\)

\(\displaystyle 4x(x + 3) = 0\)

(one linear factor) = \(\displaystyle (x + 3)\)

Roots:

\(\displaystyle x = 0\)

or

\(\displaystyle x = -3\)

Plugin those values into the original equation to see if each one is true.