Radical Problem

[Jason76]

Found this in a math book:



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

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

\(\displaystyle \sqrt{x + 4}^{2} = (3 - \sqrt{1- x})^{2}\) How would the right side be FOILED exactly?

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

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

Both sides are squared to get rid of the square root on the left side.

\(\displaystyle (6\sqrt{(1 - x)}^{2} = (6 - 2x)^{2}\) Next, the right side of this equation will be foiled. Is it the "difference of two squares" or not?

\(\displaystyle 36(1 - x) = 66 - 24x + 4x^{2}\) Mistake: 66 should be 36

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

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

\(\displaystyle x = 0 \)

\(\displaystyle x = -3\)

 

[tkhunny]

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

Before Squaring ANYTHING, think about the Domain. \(\displaystyle 1 - x \ge 0\;or\;x \le 1\) If you get anything greater than 1, throw it out.

Before squaring anything, think about simplification. '2' is wild, here.

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

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

No, there is no relationship to a "Difference of Squares". Why are you even thinking that? You are squaring a binomial.

\(\displaystyle 9(1-x) = (3 - x)^{2} = 9 - 6x + x^{2}\)

Simplify (Distributive Property)

\(\displaystyle 9 - 9x = 9 - 6x + x^{2}\)

No 66 or 36 to get wrong in there. Simplify your life up front. So much less to go wrong along the way.

 

[Jason76]

\(\displaystyle \sqrt{x + 4}^{2} = (3 - \sqrt{1- x})^{2}\) How would the right side be FOILED exactly?

Let me give it a shot:

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

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

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

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

 

[JeffM]

\(\displaystyle \sqrt{x + 4}^{2} = (3 - \sqrt{1- x})^{2}\) How would the right side be FOILED exactly?

Let me give it a shot:

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

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

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

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

Jason

(a + b)(c + d) = ac + ad + bc + bd no matter what a, b, c, and d are. It derives from general properties of numbers. A radical is just a number.