Question

I neglected to truncate the problem list when changing the page for the new edition.

However those problems on the list which are also in the text include all problems on the Query, and the numbers in the Query should match the numbers in the text.

12 = 4 * 3. The entire 4 * 3 is inside the square root, so you should have written sqrt (4 * 3), not sqrt 4 * 3; the latter would mean (sqrt 4) * 3, which is not correct.

sqrt(12) = sqrt(4 * 3) = sqrt(4) * sqrt(3)

sqrt(4) = 2 so -8 sqrt(4) = -8 * 2 = -16.

-16 sqrt(3) + sqrt(3) = (-16 + 1) sqrt(3), by the distributive law (i.e., you factor out the sqrt(3), and -16 + 1 = -15.

x/x + 4 means (x/x) + 4, which would be 1 + 4 = 5.

If you mean x / (x + 4), this is in a generally accepeted simplified form; what were the instructions for the question?

I suspect you mean either

((3x^-1)/(4y^-1))^-2 , which would be (3x^-1)^-2 / (4 y^-1)^-2 = 3^-2 * x^2 / (4^-2 * y^2) = 1/9 x^2 / (1/16 y^2) = 16/9 x^2 / y^2,

or

( (3x^(-1/4)y^(-1) )^-2 = (3 x^(-1/4))^-2 * (y^-1)^-2 = 3^-2 x^(1/2) * y^2 = 1/9 x^(1/2) y^2.

If it's neither of these, then please clarify, and also tell me what you do and do not understand about the above.

I suspect you mean

4x^2/(x^2-16) * (x^3-64)/ (2x). All these parentheses are necessary.

x^2 - 16 = (x-4)(x+4), and (x^3-64) = (x - 4) ( x^2 + 4 x + 16)

You book includes the special formulas a^3 - b^3 = (a - b) ( a^2 + a b + b^2) and a^3 + b^3 = (a + b) ( a^2 - ab + b^2).

A 3 as part of the radical sign means the 'third root', which is the 1/3 power. Also I suspect that the 81x^4y^2 is all part of the denominator and therefore needs to be grouped.

If so what you have is

(3xy^2/(81x^4y^2) )^(1/3) =

(3 (x / x^4) ( y^2 / y^2) ) ^ (1/3) =

(3 / x^3) ^ (1/3) =

3^(1/3) / (x^3)^(1/3) =

3^(1/3) / x.

sqrt(54) = sqrt( 9 * 6) = sqrt(9) * sqrt(6) = 3 sqrt(6).

The quadratic formula says that

a x^2 + b x + c = 0

if, and only if, x = (-b +- sqrt(b^2 - 4 a c) ) / (2 a) ).

When you plug in a, b and c, the quantity under the square root is either positive, negative or zero.

If it's positive, then the square root is positive, and you get two possible solutions, one for the + of the +- and one for the -.

If it's negative, then the square root is not a real number and you have no solutions.

If it's zero, then the square root is 0 and gives you the same result whether you use the + or the - of the +-. So you have only one solution.

Note that the discriminant is b^2 - 4 a c, not sqrt(b^2 - 4 a c). The solution x = (-b +- sqrt(b^2 - 4 a c) ) / (2 a) ) could also be written x = (-b +- sqrt(discriminant) / (2 a) )

You understand that | 5 | = 5 and | -5 | = 5. So

| x | = 5 is true provided either x = 5 or x = -5.

Similarly

| x - 12 | = 5 means that either x - 12 = 5 or x - 12 = -5.

If x - 12 isn't 5 or -5, the its absolute value isn't 5.

| x | < 5 is so for an infinite number of values of x, among them the integer values x = -4, -3, -2, -1, 0, 1, 2, 3, 4.

It's also true for all fractions and irrational numbers between these number, and if you think about this you will see that it is so for all values of x which lie between x = -5 and x = 5.

So | x | < 5 is the same as -5 < x < 5.

Similarly |x - 12 | < 5 means that x - 12 must lie between -5 and 5, so

| x - 12 | < 5 means -5 < x - 12 < 5.

| x | > 5 is so for an infinite number of x values, including integer values x = 6, 7, 8, 9, ... . It is also true for x = -6, -7, -8, -9, ... . In fact is it true for any x value greater than 5, and also for any x value less than 5.

So

| x | > 5 means

x > 5 OR x < -5.

Similarly

| x - 12 | < 5 means

x - 12 > 5 OR x - 12 < 5.

There is an error in that statement.

If you had instead the equation

| t | + 6 = 7, you would subtract 6 from both sides to get

| t | = 1, which is true for t = 1 or t = -1.

However the given equation

| t | + 7 = 6 yields

| t | = -1.

Since | t | can't be negative, there is no solution to this equation.

It depends on what those points represent.

If we assume that the points are on a diameter of the circle, then since the center is the midpoint of the diameter, you first find the midpoint.

The midpoint between (1, 2) and (4,2) is ( (1 + 4) / 2, (2 + 2) / 2) = (5/2, 2).

The radius is the distance from the center to the end of a diameter, and it is also half the diameter. So you have a choice.

Let's choose to use half the diameter.

The diameter is the distance from (1, 2) to (4, 2), which is sqrt( (4 - 1)^2 + (2 - 2) )^2 = sqrt(3^2 + 0^2) = sqrt( 9 + 0) = sqrt(9). = 3.

So the radius is 1/2 * 3 = 3/2.

Had we calculated the distance from (5/2, 2) to either of the two points, the result would have been identical.

The same sequence of steps would work for the second set of point.

sqrt(12) = sqrt( 4 * 3) = sqrt(4) * sqrt(3) = 2 * sqrt(3).

Good, but 6 divided by 3 is 2, not 3, so you should have

(x+1)^2 + (y-1)^2 = 2.

Looking below it looks like that's what you have;' probably just a typo.

You're doing fine up to this point.

y^2 - 2y = 0 is a quadratic equation with a = 1, b = -2 and c = 0, so you could use the quadratic formula. You would find that y = 0 or y = 2.

If you can factor a quadratic equation, it's generally easier to solve using factoring than using the formula. In this case the equation

y^2 - 2y = 0 factors as

y ( y - 2) = 0. The product of two quantities can only be 0 if one of the quantities is 0, so the solution is

y = 0 or y-2 = 0, which gives you

y = 0 or y = 2,

which agrees with the result of the quadratic formula.

Good questions. In some of the earlier questions you didn't use appropriate grouping, so it wasn't possible to tell what the question really was. I tried to answer all reasonably possible interpretations; however you need to be sure you understand the need for grouping and use it appropriately.