Hot X: Algebra Exposed
Solution Guide for Chapter 24
Here are the solutions for the “Doing the Math” exercises in Hot X: Algebra Exposed!
DTM from p.356
2. Let’s find the values of x that make each parentheses equal to zero! We’ll solve
x + 2 = 0 by subtracting 2 from both sides and we get x = –2. Similarly, x – 1 = 0 will be
true when x = 1. So those are the values of x that satisfy the equation. For the next part,
we’ll multiply out the parentheses to see what this equation looked like before it was
factored:
(x + 2)(x – 1) = 0

x2 – x + 2x – 2 = 0

x2 + x – 2 = 0
Done!
Answer: x2 + x – 2 = 0 is solved by x = –2, 1.
3. Let’s find the values of y that make each parentheses equal to zero! Well, they’ll both
be zero if y = –2; notice that there’s no other number that would make that happen. So
that’s our solution. For the next part, we’ll multiply out the parentheses to see what this
equation looked like before it was factored:
(y + 2)(y + 2) = 0

y2 + 2y + 2y + 4 = 0

y2 + 4y + 4 = 0
Answer: y2 + 4y + 4 = 0 is solved by y = –2.
4. To find the value of t that makes the first parentheses equal to zero, we need to solve
the equation 3t – 1 = 0. We just add 1 to both sides and then divide by 3, and we get:
t=
1
. Similarly, t – 4 = 0 will be true when t = 4. Now let’s multiply out the
3
parentheses, and we get: (3t – 1)(t – 4) = 0  3t2 – 12t – t + 4 = 0 
3t2 – 13t + 4 = 0.
Answer: 3t2 – 13t + 4 = 0 is solved by t =
1
, 4.
3
5. To find the value of x that makes the first parentheses equal to zero, we need to solve
the equation 2x + 3 = 0. We just subtract 3 from both sides and then divide by 2, and we
get: x = –
3
5
. Similarly, 3x – 5 = 0 will be true when x = . Now let’s multiply out the
2
3
parentheses, and we get
(2x + 3)(3x – 5) = 0  6x2 – 10x + 9x – 15 = 0  6x2 – x – 15 = 0
Answer: 6x2 – x – 15 = 0 is solved by x = –
3 5
,
2 3
DTM from p.360
2. Factoring this, we can set up the parentheses: (x )(x ) and we need two numbers
whose product is –7 and whose sum is 6, which is just 7 & –1! So the equation factors
into (x + 7)(x – 1) = 0. So this will be true when x = –7 or x = 1. Let’s check these
solutions. First, we’ll plug in x = –7, and we get:
x2 + 6x – 7 = 0
 (–7)2 + 6(–7) – 7 = 0 ?
 49 – 42 – 7 = 0 ?
 7 – 7 = 0 ? Yep!
Checking the other solution, x = 1, we get:
x2 + 6x – 7 = 0
 (1)2 + 6(1) – 7 = 0 ?
1+6–7=0?
 7 – 7 = 0? Yep!
Answer: x = –7, x = 1
3. In order to factor this, we should first get zero on one side, so we subtract 7 from both
sides, and we get: x2 + 6x – 7 = 0. Hey, that’s the same problem as #2, so we know the
answer!
Answer: x = –7, x = 1
4. In order to factor this, let’s first get zero on one side, so let’s subtract 5w from both
sides and also subtract 5 from both sides, and we get:
10w2 – 5w – 5 = 0. We can first factor out 5 from every term and we get:
5(2w2 – w – 1) = 0. We can divide both sides by 5, and of course, zero divided by 5 is
just zero! So our equation is equivalent to: 2w2 – w – 1 = 0, which we can then factor like
this: (2w + 1)(w – 1) = 0, using our techniques from chapter 22. And according to the
Zero-Product property from p. 354, this is true when
w =–
1
or w = 1.
2
Let’s check this answer!
1
Checking w = − , we get:
2
10w2 – 5w – 5 = 0
1
1
 10( − )2 – 5( − ) – 5 = 0 ?
2
2
 10(
1
5
) + – 5 = 0 ? let’s multiply both sides by 4, to make this easier!
4
2
And we get:
 10 +
20
– 20 = 0 ?
2
 10 + 10 – 20 = ?  20 – 20 = 0? Yep!
Let’s check the other solution, w = 1:
10w2 – 5w – 5 = 0
 10(1)2 – 5(1) – 5 = 0 ?
 10 – 5 – 5 = 0?  5 – 5 = 0? Yep!
Answer: w = –
1
,1
2
5. In order to factor this, let’s first get zero on one side, so let’s subtract 20y from both
sides, and we get: 4y2 – 20y – 24 = 0. We can first factor out 4 from every term and we
get: 4(y2 – 5y – 6) = 0. We can divide both sides by 4, and of course, zero divided by 4
is just zero! So our equation is equivalent to: y2 – 5y – 6 = 0, which we then factor like
this: (y – 6)(y + 1) = 0. This will be true when y = 6 or –1. Let’s check these solutions.
First we’ll start with y = 6:
4y2 – 20y – 24 = 0
 4(6)2 – 20(6) – 24 = 0 ?
 4(36) – 120 – 24 = 0 ?
 144 – 120 – 24 – 0 ?
 24 – 24 = 0? Yep!
Next we’ll check the solution y = –1:
4y2 – 20y – 24 = 0
 4(–1)2 – 20(–1) – 24 = 0 ?
 4 + 20 – 24 = 0 ?
 24 – 24 = 0? Yep!
Answer: y = –1, 6
6. This equation looks like it’s already factored, but we won’t be fooled! Since zero isn’t
on one side, we need to multiply this out, then get zero on one side, and then factor it.
Let’s do it! Multiplying out the left side, this equation becomes:
(x – 2)(x + 3) = 14 
x2 + 3x – 2x – 6 = 14  x2 + x – 20 = 0. Great, now it’s ready
to be factored like we did in chapter 22, which is: (x + 5)(x – 4) = 0. This will be true
when x = –5 or 4. Let’s make sure these solutions are right! First we’ll check to see what
happens when x = –5:
(x – 2)(x + 3) = 14
 (–5 – 2)(–5 + 3) = 14 ?
(–7)(–2) = 14?  14 = 14? Yep!
Now let’s check to see if the solution x = 4 makes the equation true:
(x – 2)(x + 3) = 14
 (4 – 2)(4 + 3) = 14 ?
(2)(7) = 14?  14 = 14? Yep!
Answer: x = –5, 4
7. This will be much easier without all those negatives; let’s start by multiplying both
sides by –1, and we get: 4h2 + 3h = 1. Now we’ll subtract 1 from both sides, and we get:
4h2 + 3h – 1 = 0. Let’s factor this like we did in chapter 22, and we get:
(h + 1)(4h – 1) = 0. This will be true when h = –1 or h =
Let’s check these; first h = –1:
–4h2 – 3h = –1
 –4(–1)2 – 3(–1) = –1 ?
 –4 + 3 = –1 ?
 –1 = –1 ? yep!
Now we’ll check the other solution, h =
–4h2 – 3h = –1
 –4(

1 2
1
) – 3( ) = –1 ?
4
4
−4 3
− = –1 ?
16 4
−
1 3
− = –1?  –1 = –1 ? Yep!
4 4
Answer: y = –1,
1
4
1
:
4
1
.
4
DTM from p.363-364
2. Ok, so we’re supposed to build an equation that has the solutions x = –1 and x = 5.
Let’s figure out two parentheses that can multiply together, each equaling zero for one of
the solutions. Well, x = –1 is the solution to (x + 1) = 0, right? Also, x = 5 is the solution
to (x – 5) = 0. That means x = –1 and x = 5these are the solutions to the big equation:
(x + 1)(x – 5) = 0. Great! Now let’s multiply this out, and we get: x2 – 5x + x – 5 = 0 
x2 – 4x – 5 = 0. And that’s the equation that has the solutions x = –1 and x = 5. If you
plug either of those solutions in for x, you’ll get a true statement – try it!
Answer: x2 – 4x – 5 = 0
3. Let’s build an equation that has the solutions x = 1 and x = –5. Well, x = 1 is the
solution to the equation (x – 1) = 0, and x = –5 is the solution to (x + 5) = 0. This means
these are the two solutions to the quadratic equation: (x – 1)(x + 5) = 0. Let’s multiply
this out and we get:
(x – 1)(x + 5) = 0  x2 + 5x – x – 5 = 0  x2 + 4x – 5 = 0
And that’s the equation that has the solutions x = 1 and x = –5. Very similar to the above
problem, huh!
Answer: x2 + 4x – 5 = 0
4. Okay, building an equation that has the solutions x = 1 and x = 5, we know that x = 1 is
the solution to the equation (x – 1) = 0, and x = 5 is the solution to (x – 5) = 0, right?
So then x = 1 and x = 5 are the solutions to the quadratic equation: (x – 1)(x – 5) = 0.
Let’s multiply this out and we get:
(x – 1)(x – 5) = 0  x2 – 5x – x – 5 = 0  x2 – 6x – 5 = 0
And that’s the equation that has the solutions x = 1 and x = 5. Nice!
Answer: x2 – 6x – 5 = 0
5. Ok, we’re supposed to build an equation that has the solutions x = –1 and x = –5. Let’s
figure out two parentheses that can multiply together, each equaling zero for one of the
solutions. Well, x = –1 is the solution to (x + 1) = 0, right? Also, x = –5 is the solution to
(x + 5) = 0. That means these are the two solutions to the big equation: (x + 1)(x + 5) = 0.
Great! Now let’s multiply this out, and we get: x2 + 5x + x + 5 = 0  x2 + 6x + 5 = 0.
And that’s the equation that has the solutions x = –1 and x = –5.
Answer: x2 + 6x + 5 = 0
6. Let’s build an equation that has the solutions x =
solution to the equation (x –
2
1
2
and x = – . Well, x = is the
3
5
3
2
) = 0, but let’s multiply both sides by 3 to get rid of the
3
fraction (which will make life easier when we go to multiply our two parentheses
together!), and we find that 3(x –
solution x =
2
) = 3(0)  (3x – 2) = 0 is also an equation with the
3
2
1
. With me so far? Similarly, x = – is the solution to the equation
3
5
(5x + 1) = 0.
This means x =
2
1
and x = – are the solutions to the quadratic equation:
3
5
(3x – 2)(5x + 1) = 0. Let’s multiply this out and we get:
(3x – 2)(5x + 1) = 0  15x2 + 3x – 10x – 2 = 0  15x2 – 7x – 2 = 0
And that’s the equation that has the solutions x =
2
1
and x = – . Great!
3
5
Answer: 15x2 – 7x – 2 = 0
7. It’s up to you! (C’mon, go ahead and do one…)