Question

This problem addresses some common algebraic errors. For the equalities stated below assume that x and y stand for real numbers. Assume that any denominators are non-zero. Mark the equalities with T (true) if they are true for all values of x and y, and F (false) otherwise.(x+y)^2 = x^2 + y^2.(x+y)^2 = x^2 + 2xy+y^2.x/ (x+y)= 1/y.x-(x+y) = y.sqrt{x^2} = x.sqrt{x^2} = |x|.sqrt{x^2+4} = x+2.1(x+y) = 1/x + 1/y.

Answer 1

Answer:

The answers are:

a) $(x+y)^2 = x^2 + y^2$ is FALSE.

b) $(x+y)^2 = x^2 + 2xy + y^2$ is TRUE.

c) $\frac{x}{x+y}= \frac{1}{y}$ is FALSE.

d) $x-(x+y) = y$ is FALSE.

e) $\sqrt{x^2} = x$ is FALSE.

f) $\sqrt{x^2} = |x|$ is TRUE.

g) $\sqrt{x^2+4} = x+2$ is FALSE.

h) $\frac{1}{x+y} = \frac{1}{x} + \frac{1}{y}$ is FALSE.

Step-by-step explanation:

We can show that the FALSE statement are, in fact, false finding an example where the equality fails.

a) Take x=2 and y=3 and notice that (2+3)² = 25, while 2²+3²=4+9=13.

b) (x+y)² = (x+y)(x+y)=x²+xy+xy+y² = x² +2xy +y².

c) Take x=2 and y=3 and notice that 2/(2+3) = 2/5 which is different from 1/3.

d) Take x=2 and y=3 and notice that 2-(2+3) = 2-5=-3 which is different from 3. Recall that x-(x+y) = x-x-y=-y.

e) Take x=-2, and notice that [tax]\sqrt{(-2)^2} = \sqrt{4} = 2[/tex] which is different from -2.

f) The previous example illustrate why this is true.

g) Take x=1, and notice that $sqrt{1^2+4} = \sqrt{5}$ which is different from 3.

h) Take x=2 and y=3 and notice that 1/(2+3)=1/5 and 1/3+1/2 = 5/6.