## 1.

*x*:

## 2.

To factor a quadratic where the second-order term has a coefficient of 1, find two numbers that multiply up to the constant, –7, and add up to the first-order term’s coefficient, 6. These two numbers are 7 and –1, so you can rewrite the quadratic equation as follows:

*x*

^{2}+ 6

*x*– 7 = (

*x*+ 7)(

*x*– 1)

To find the solutions—or roots of the resulting quadratic function—set the equation equal to zero. You can see that this happens when *x* = –7 or *x* = 1. (F) is the only answer choice that lists one of these options.

If you chose (H) or (J), you probably forgot that a root is the negative of one of those two numbers you found, because it’s a value of *x* that makes one of the bracketed expressions equal to zero.

## 3.

^{st}term, thankfully, you can use the formula for an arithmetic sequence,

*a*=

_{n}*a*

_{1}+

*d*(

*n*– 1) , to find the

*n*th term algebraically. Simply set

*a*

_{1}= 11,

*d*= –3, and

*n*= 21 for the 21

^{st}term in a sequence where the first term is 11 and it decreases by 3 each term, and solve:

If you got any of the other answer choices, you found a term from the right sequence, but not the 21^{st} term.

## 4.

*r*and

*q*in terms of

*x*, so you can write out the value of

*y*also in terms of

*x*:

*x*to see how the value of

*y*changes with respect to

*x*:

*x*is halved,

*y*decreases by a factor of .

## 5.

Taking the second inequality, *x* > 1, into account, you know that the number line needs to show values between 1 and 4, including 4 and excluding 1. The only answer choice representing this is (E).

## 6.

*a*

*x*

^{2}+

*bx*+

*c*, you need to find two values that add to

*b*and multiply to

*ac*. In this case, two numbers that multiply to 2 × (–1) = –2 and add to –1 are –2 and 1. Split the middle term so that these numbers are the resulting coefficients, then factor as usual:

You can eliminate (G) because 2*x* × 2*x* does not equal 2*x*^{2}, *a*.

You can eliminate (J) and (K) because neither (–1) × (–1) nor 1 × 1 equals –1, *c*.

You can eliminate (F) because (1 × (–1)) + (1 × 2) = –1 + 2 = 1, and *b* should equal –1.

## 7.

*x*:

*x*are –1 and 6. (A) is the only choice that lists one of these solutions.

## 8.

*y*=

*x*+ 1 to plug in the value for

*y*into the second equation:

*x*:

You can substitute this back into the equations to confirm your answer.

## 9.

Note that this is a bit tricky, because you have to see that I is possible even while the volume in the tank is increasing. The volume is not increasing as much as before, so I may be the result of this.

## 10.

## 11.

*b*(

*w*) is the number of blocks of pressed boards that the manager wishes to produce:

## 12.

This means that after 5 hours, (8 boards per hour)(5 hours) = 40 boards.

Next, find the number of boards that would have been produced with the old method:

The difference between the two methods is 40 – 10 = 30 boards.

## 13.

^{x}= 128

*x*, you will find that

*x*= 7. You can confirm this by quickly plugging 2

^{7}into your calculator.

## 14.

If you got any other result, check your algebra as it is most likely due to incorrect simplifying and elimination.

## 15.

*i*raised to a power repeats itself in cycles of 4.

*i*

^{1}=

*i*,

*i*

^{2}= –1,

*i*

^{3}= –1,

*i*

^{4}= 1… To find the value of

*i*

^{356}, divide the exponent by 4: . Since there is no remainder,

*i*

^{356}is equivalent to

*i*

^{4}= 1.

## 16.

On the second day, Lucy collects 4 cans; on the third, 8; on the fourth, 16; on the fifth, 32. Adding all the cans together gives 2 + 4 + 8 + 16 + 32 = 62 cans in total.

If you chose (H), you probably picked the number of cans collected on the fifth day alone, rather than the cumulative number of cans collected.

## 17.

## 18.

*V*=

*F*, and

*E*= 6:

Substituting *V* into the first equation:

## 19.

*n*>

*m*, then must be greater than 1. Use the rule that any logarithm in the form . However, must be greater than 6 since > 1. This means that . You know then that

*p*> 1.

If you chose another answer, make sure to review your log rules.

## 20.

*i*is in the numerator and not the denominator. You do this by multiplying both the numerator and denominator by the value

*i*. This doesn’t change the value of the expression, but will shift the

*i*to the numerator: