**Problem 1**

For which function defined by a polynomial are the zeros of the polynomial −4 and −6?

1) *y *= *x*^{2} − 10*x *− 24

2) *y *= *x*^{2} + 10*x *+ 24

3) *y *= *x*^{2} + 10*x *− 24

4) *y *= *x*^{2} − 10*x *+ 24

*Discussion and Solution:*

There are two ways of solving this problem – (1) solve each equation for the value of x and (2) substitute each possible solution into each equation to which one works. Let’s try them both.

Method (1) –

*y*=*x*^{2}– 10*x*– 24

0 = *x*^{2} – 10*x* – 24

0 = (*x* – 4)(*x* – 6)

0 = (*x* – 4); *x* = 4

0 = (*x* – 6); *x* = 6

No way

*y*=*x*^{2}+ 10*x*+ 24

0 = *x*^{2} + 10*x* + 24

0 = (*x* + 4)(*x* + 6)

0 = (*x* + 4); *x* = – 4

0 = (*x* + 6); *x* = – 6

Bingo – they work here – we can stop here since we have the answer

Method (2) –

*y*=*x*^{2}– 10*x*– 24

0 = *x*^{2} – 10*x* – 24

0 = (– 4)^{2} – 10(– 4) – 24

0 = 16 + 40 – 24

0 ≠ 32

No way

*y*=*x*^{2}+ 10*x*+ 24

0 = *x*^{2} + 10*x* + 24

0 = (– 4)^{2} + 10(– 4) + 24

0 = 16 – 40 + 24

0 = 0

This one works, but we better check the other one.

*y* = *x*^{2} + 10*x* + 24

0 = *x*^{2} + 10*x* + 24

0 = (– 6)^{2} + 10(– 6) + 24

0 = 36 – 60 + 24

Bingo. This one works also.

Correct solution is (2).

**Problem 2**

If *f*(*x*) = *x*^{2} − 2*x *− 8 and *g*(*x*) = ¼ *x *− 1, for which values of *x *is *f*(*x*) = *g*(*x*)?

1) −1.75 and −1.438

2) −1.75 and 4

3) −1.438 and 0

4) 4 and 0

*Discussion and Solution:*

The easiest way to solve this is to substitute each possible solution into each equation and see which one clicks.

Since options (3) and (4) both have 0 as a possible solution, let’s start with this one and see what happens.

*f*(*x*) = *x*^{2} − 2*x *– 8 = (0)^{2} – 2(0) – 8 = – 8

*g*(*x*) = ¼ *x *– 1 = ¼ (0) – 1 = – 1

I don’t know about you, but I was taught – 8 ≠ – 1. Therefore, choices (3) and (4) are out.

Now let’s try options (1) and (2). Let’s try the 4.

*f*(*x*) = *x*^{2} − 2*x *– 8 = (4)^{2} – 2(4) – 8 = 16 – 8 – 8 = 0

*g*(*x*) = ¼ *x *– 1 = ¼ (4) – 1 = 1 – 1 = 0

* *That works, now let’s try the – 1.75.

*f*(*x*) = *x*^{2} − 2*x *– 8 = (– 1.75)^{2} – 2(– 1.75) – 8 = 3.0625 + 3.5 – 8 = – 1.4375

*g*(*x*) = ¼ *x *– 1 = ¼ (– 1.75) – 1 = – 0.4375 – 1 = – 1.4375

That’s okay, but let’s try the – 1.438 to rule out choice (1).

*f*(*x*) = *x*^{2} − 2*x *– 8 = (– 1.438)^{2} – 2(– 1.438) – 8 = 2.0678 + 2.876 – 8 = – 3.056

*g*(*x*) = ¼ *x *– 1 = ¼ (– 1.438) – 1 = – 0.3595 – 1 = – 1.3595

Nope.

Choice (2) is the correct choice.

**Problem 3**

Which system of equations has the same solution as the system below?

2*x *+ 2*y *= 16

3*x *− *y *= 4

1) 2*x *+ 2*y *= 16

6*x *− 2*y *= 4

2) 2*x *+ 2*y *= 16

6*x *− 2*y *= 8

3) *x *+ *y *= 16

3*x *− *y *= 4

4) 6*x *+ 6*y *= 48

6*x *+ 2*y *= 8

* **Discussion and Solution:*

In order to have the same solution, when we divide the possible solutions by some small number, say a 2 or 3, we should arrive at the original equation. Here goes.

Choice (1):

1) 2*x *+ 2*y *= 16

6*x *− 2*y *= 4

The first equation is the same as one of the original equations. As for the second equation, the left side appears to be doubled from the original equation, but not the right side. This choice is out.

Choice (2):

2) 2*x *+ 2*y *= 16

6*x *− 2*y *= 8

The first equation is the same as one of the original equations. As for the second equation, both sides have been doubled as compared to the other original equation. This choice is the correct one.

Choice (3):

3) *x *+ *y *= 16

3*x *− *y *= 4

In this choice, the second equation is the same as one of the original equations, but the first equation is messed up. The right side of this equation is the same as the right side of the other original equation, but the left side has been halved. This choice is out.

Choice (4):

4) 6*x *+ 6*y *= 48

6*x *+ 2*y *= 8

In this choice, the first equation is the triple of one of the original equations, but the second one is messed up. At first glance, it appears that the equation has merely been doubled. Take a closer look. If it were just doubled, it would be 6*x* – 2*y* = 8, not 6*x* + 2*y* = 8. This choice is out.

Choice (2) is the correct one.

**Problem 4**

Mo’s farm stand sold a total of 165 pounds of apples and peaches. She sold apples for $1.75 per pound and peaches for $2.50 per pound. If she made $337.50, how many pounds of peaches did she sell?

1) 11

2) 18

3) 65

4) 100

* **Discussion and Solution:*

Let’s set up two equations and solve. *a* = number pounds of apples. *p* = number of pounds of peaches.

*a* + *p* = 165

1.75*a* + 2.50*p* = 337.50

Two equations with two unknowns. Don’t you just love that? As you’re probably aware, when we have two equations with two unknowns, there are three ways of solving this. They are:

(a) Solve one equation for one of the unknowns and substitute this into the other equation and solve.

(b) Graph the two equations and look to see where they intersect. This point of intersection will indicate the solutions. If the lines are parallel or on top of each other, then there is no common solution.

(c) Multiply one of the equations by some number (in this case, multiply the top equation by 1.75) and subtract the two equations, thereby eliminating one of the variables. Solve for the remaining variable. Now, substitute this value into either of the original equations and determine the missing value.

Let’s use procedure (a).

*a* + *p* = 165

a = 165 – p

1.75*a* + 2.50*p* = 337.50

1.75(165 – p) + 2.50p = 337.50

288.75 – 1.75p + 2.50p = 337.50

288.75 + 0.75p = 337.50

0.75p = 337.50 – 288.75 = 48.75

*p* = 65

Choice (3) is the correct choice.

**Problem 5**

If *A *= 3*x*^{2} + 5*x *− 6 and *B *= −2*x*^{2} − 6*x *+ 7, then *A *− *B *equals

1) −5*x*^{2} − 11*x *+ 13

2) 5*x*^{2} + 11*x *− 13

3) −5*x*^{2} − *x *+ 1

4) 5*x*^{2} − *x *+ 1

* **Discussion and Solution:*

Since we’re looking for *A* – *B*, we merely need to set the two equations equal to each other.

*A* – *B*

(3*x*^{2} + 5*x *– 6) – (−2*x*^{2} − 6*x *+ 7)

3*x*^{2} + 5*x *– 6 + 2*x*^{2} − 6*x – *7

5*x*^{2} + 11*x* – 13

Choice (2) is correct.