Solving a Two Radical Equation for Roots

**Jason76** · October 19th 2012, 04:15 AM

How do I get the roots from this equation?

$\sqrt {x + 4} + \sqrt{1 - x} = 3$

**TheEmptySet** · October 19th 2012, 04:24 AM

Originally Posted by Jason76

How do I get the roots from this equation?

$\sqrt {x + 4} + \sqrt{1 - x} = 3$

1. Isolate a radical

2. Square both sides

3. If the equation contains a radical go back to step1, if not move on to step 4

4. Solve the equation

5. check your solutions. extraneous solutions may have been introduced when the equation was squared.

$\sqrt{x+4}=3-\sqrt{1-x}$

$(\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)$

$-6 = -6\sqrt{1-x} \iff 1=\sqrt{1-x}$

$1=1-x \iff x=0$

**Jason76** · October 19th 2012, 04:33 AM

I basically get the jist of it but I don't understand the following line:

$(\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)$

So

$\sqrt{x+4}$

was squared (to get rid of the square root), so now the same has to be done to the other side. However,

$3$ (to the right of the equation)

was not under square root. So why was it squared?

Obviously,

$\sqrt{1-x}$

was squared to get rid of the radical sign.

**TheEmptySet** · October 19th 2012, 04:41 AM

Originally Posted by Jason76

I basically get the jist of it but I don't understand the following line:

$(\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)$

So

$x + 4$

which was square rooted was squared (to get rid of the square root), so now the same has to be done to the other side. However,

$- 3$

was not under square root. So why was it squared?

If you multiply two binomials together you do not get

$(a+b)^2 \ne a^2+b^2$

You must use the distributive property to get

$(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$

**Jason76** · October 19th 2012, 04:58 AM

Originally Posted by TheEmptySet

If you multiply two binomials together you do not get

$(a+b)^2 \ne a^2+b^2$

You must use the distributive property to get

$(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$

Considering again:

$(\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)$

Right, but

$\sqrt{x+4}$

is a binomial, yet, after squaring, no distributive property is used on it. On the other hand,

$- \sqrt{1-x}$

(after squaring) gets a distributive property operation done to it. Plus, I don't understand where the

$9 - 6$

really came from.

Again:

$(\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)$

Anyhow, what I'm trying to say is "Two binomials exist on both sides of the equation. Both of them should be squared (to get rid of the radical signs). Therefore, the binomial on the left is squared, as expected. However, the one on the right is different. It is squared and then distributed. Yet, the radical sign remains for one of the duplicate terms.". Plus, I still have no idea where the

$9 - 6$

came from. I'm assuming the $9$ came from squaring.

**Jason76** · October 19th 2012, 06:43 PM

Sorry to bump thread, but I couldn't edit the last post.

Here is what I'm saying more clearly:

How do I solve this for roots?

$\sqrt{x+4} + \sqrt{1-x} = 3$

Put one radical on the right side of the equation.

$\sqrt{x + 4} = 3 - \sqrt{1 - x}$

Square both sides to get rid of radicals.

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

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

$x + 4 = 9 - 1 - x$

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

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

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.

**MarkFL** · October 19th 2012, 06:53 PM

Look at the rule TheEmptySet gave in post #4...this will tell you why the squaring of the side in question becomes what it does.

**Jason76** · October 20th 2012, 06:22 AM

I think I got it now:

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

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

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

FOIL

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

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

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

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

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

Full equation:

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

Next, more algebra until the roots are found.

**Jason76** · October 20th 2012, 07:57 PM

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

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

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

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

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

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

$4x^2 + 12x = 0$

$4x(x + 3) = 0$

(one linear factor): $(x + 3)$

Roots:

$x = 0$

or

$x = -3$

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

One question. Where did the 0 come from?

**earboth** · October 20th 2012, 10:14 PM

Originally Posted by Jason76

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

...

$4x(x + 3) = 0$

(one linear factor): $(x + 3)$

Roots:

$x = 0$

or

$x = -3$

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

One question. Where did the 0 come from?

$\underbrace{4x}_{\text{1st factor}}\overbrace{(x + 3)}^{\text{2nd factor}} = 0$

A product equals zero if one of the factors equals zero. Therefore:

$4x = 0 ~\vee~x+3=0~\implies~\boxed{x=0~\vee~x=-3}$

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