PDF Hot X: Algebra Exposed - Danica McKellar

Hot X: Algebra Exposed

Solution Guide for Chapter 25

Here are the solutions for the "Doing the Math" exercises in Hot X: Algebra Exposed!

DTM from p.371

2. To solve for y, first we take the square root of both sides, and we get: (y - 1)2 = 16 (y ? 1) = ? 4. This becomes two mini-equations: y ? 1 = 4 and y ? 1 = ?4, and solving these, we get y = 5 and y = ?3. Next we'll multiply out the original equation and put it in standard form: (y ? 1)2 = 16 y2 ? y ? y + 1 = 16 y2 ? 2y + 1 = 16 y2 ? 2y ?15 = 0. Done! Answer: y2 ? 2y ?15 = 0 has the solutions y = ?3, 5

3. To solve for t, first we take the square root of both sides, and we get: (t + 3)2 = 4 (t + 3) = ? 2. This becomes two mini-equations: t + 3 = 2 and t + 3 = ?2, and solving these, we get t = ?1 and t = ?5. Next we'll multiply out the original equation and put it in standard form: (t + 3)2 = 4 t2 + 3t + 3t + 9 = 4 t2 + 6t + 9 = 4 t2 + 6t + 5 = 0 Done! Answer: t2 + 6t + 5 = 0 has the solutions t = ?5, ?1

4. To solve for x, first we take the square root of both sides, and we get: (x - 3)2 = 5 . (x ? 3) = ? 5 . This becomes two mini-equations: x ? 3 = 5 and x ? 3 = ? 5 , and solving these, we get y = 3 + 5 and y = 3 ? 5 . Next we'll multiply out the original equation and put it in standard form: (x ? 3)2 = 5 x2 ? 3x ? 3x + 9 = 5 x2 ? 6x + 9 = 5 x2 ? 6x + 4 = 0 Done! Answer: x2 ? 6x + 4 = 0 has the solutions x = 3 ? 5 , 3 + 5

DTM from p.378

2. So, x2's coefficient is 1, all the x stuff is on one side and the constant is on the other. That means we're ready to leave blanks for the missing "magic" number: x2 + 8x + ___ = 1 + ___ . That missing number will be 8 divided by 2, which is 4 (that will go in the parentheses) and then squared, which is 16. So we get: x2 + 8x + 16 = 1 + 16 And now we know we can write the left hand side as a squared binomial: (x + 4)2 = 17 But did we do this right? Let's multiply it out, just to make sure: (x + 4)2 = 17 x2 + 4x + 4x + 16 = 17 x2 + 8x + 16 = 17 x2 + 8x = 1 Yep, we got what we started with, so we did it correctly! Answer: (x + 4)2 = 17

3. First we should notice that we can divide both sides by 3, and we get: x2 + 2x ? 2 = 0. Then adding 2 to both sides, we get: x2 + 2x = 2, and we're ready to find the missing number needed to rewrite the left side as a squared binomial. x2 + 2x + ___ = 2 + ___ . That missing number will be 2 divided by 2, which is 1 (that will go in the parentheses) and then squared, which is 1. So we get: x2 + 2x + 1 = 2 + 1

And now we know we can write the left hand side as a squared binomial: (x + 1)2 = 3

But did we do this right? Let's multiply it out, just to make sure: (x + 1)2 = 3 x2 + x + x + 1 = 3 x2 + 2x + 1 = 3 x2 + 2x = 2. And if we wanted, we could multiply both sides by 3 to get 3x2 + 6x = 6 and then subtract both sides by 6 to get: 3x2 + 6x ? 6 = 0, which is what we started with, so we did it correctly! Answer: (x + 1)2 = 3

4. Let's add 3 to both sides and we get: x2 ? 6x = 3. Ready for the missing number: x2 ? 6x + ___ = 3 + ___, and we know that will be ?6 divided by 2, which is ?3 (that will go in the parentheses), and then squared, which is 9. So x2 ? 6x + 9 = 3 + 9 (x ? 3)2 = 12. And if you multiplied that out, you could get the x2 ? 6x = 3 that we started out with, so it's indeed an equivalent equation! Answer: (x ? 3)2 = 12

5. So, x2's coefficient is 1, all the x stuff is on one side and the constant is on the other.

That means we're ready to leave blanks for the missing "magic" number:

x2 + x + ___ = 2 + ___ . That missing number will be that invisible 1 divided by 2, which

is 1 (that will go in the parentheses) and then squared, which is 1 .

2

4

So we get: x2 + x + 1 = 2 + 1

4

4

x2 + x + 1 = 9 4 4

And now we know we can write the left hand side as a squared binomial: (x + 1 )2 = 9

24 And if you multiplied that out, you could get the x2 + x = 2 that we started out with, so

it's indeed an equivalent equation. Go ahead and try it! Answer: (x + 1 )2 = 9

24

DTM from p.380

2. First we'll complete the square. There's no GCF to pull out, so we can go ahead and put it in the right form, by subtracting 9 from both sides and setting it up to find the missing number we should add to both sides: x2 + 10x + __ = ?9 + ___ to find it, we take 10, divide it in half (so 5 will go in the parentheses) and then square it, which is 25. This is the "d2." So we can rewrite our equation, by adding that to both sides: x2 + 10x + 25 = ?9 + 25 x2 + 10x + 25 = 16 (x + 5)2 = 16. Yay! We figured out how to write the original equation as a squared thing on the left, and a number on the right: We've completed the square. We could check this work, but we can also just wait till the end and check our final solutions.

Now, to solve for x, we just take the square root of both sides, and we get: (x + 5)2 = 16

(x + 5) = ? 4 ...which becomes two equations: x + 5 = 4 and x + 5 = ?4, and we solve these to get the solutions x = ?1 and x = ?9. Let's check these in the original equation, first checking x = ?1: x2 + 10x + 9 = 0 (?1)2 + 10(?1) + 9 = 0 ? 1 ? 10 + 9 = 0 ? ?9 + 9 = 0 ? yep! And checking x = ?9: x2 + 10x + 9 = 0 (?9)2 + 10(?9) + 9 = 0 ? 81 ? 90 + 9 = 0 ? ?9 + 9 = 0 ? yep! Answer: d2 = 25; x = ?9, ?1

3. First we'll complete the square. There's no GCF to pull out, and it's already in the right form, so let's set it up to find the missing number we should add to both sides: x2 + 2x + __ = 4 + ___ to find it, we take 2, divide it in half (so 1 will go in the parentheses) and then square it, which is 1. This is the "d2." So we can rewrite our equation, by adding that to both sides: x2 + 2x + 1 = 4 + 1 x2 + 2x + 1 = 5 (x + 1)2 = 5. Yay! We figured out how to write the original equation as a squared thing on the left, and a number on the right: We've completed the square. And it's easy to multiply this out and check our work to see how we can get what we started with.

Now, to solve for x, we just take the square root of both sides, and we get: (x + 1)2 = 5

(x + 1) = ? 5 ...which becomes two equations: x + 1 = 5 and x + 1 = ? 5 , and we solve these to get the solutions x = ?1 + 5 and x = ?1 ? 5 . These are irrational solutions, so we don't really need to worry about checking them, it's a bit advanced for this stage of the game. Done! Answer: d2 = 1; x = ?1 ? 5 , ?1 + 5

4. We can factor out a GCF of 3 (in other words, divide both sides by 3), and we get: x2 ? 4x = 2, and it's already in the right form, so let's set it up to find the missing number we should add to both sides: x2 ? 4x + __ = 2 + ___ to find it, we take ?4, divide it in half (so ?2 will go in the parentheses) and then square it, which is 4. This is the "d2." So we can rewrite our equation, by adding 4 to both sides: x2 ? 4x + 4 = 2 + 4 x2 ? 4x + 4 = 6 (x ? 2)2 = 6. Yay! We figured out how to write the original equation as a squared thing on the left, and a number on the right: We've completed the square. And it's easy to multiply this out and check our work to see how we can get what we started with. Try it!

Now, to solve for x, we just take the square root of both sides, and we get: (x - 2)2 = 6

(x ? 2) = ? 6 ...which becomes two equations: x ? 2 = 6 and x ? 2 = ? 6 , and we solve these to get the solutions x = 2 + 6 and x = 2 ? 6 . These are irrational solutions, so we don't really need to worry about checking them, you'll do more of that in Algebra II. Done! Answer: d2 = 4; x = 2 ? 6 , 2 + 6

5. There's no GCF to pull out, and it's already in the right form, so we can go ahead and

put it in the right form, by adding 4 to both sides and setting it up to find the missing

number:

x2 + 3x + __ = 4 + ___

3 to find it, we take 3, divide it in half (so will go in the parentheses) and then square it,

2

which is 9 . This is the "d2." So we can rewrite our equation, by adding that to both 4

sides: x2 + 3x + 9 = 4 + 9 x2 + 2x + 9 = 16 + 9 (x + 3 )2 = 25 .

4

4

4 4 4

24

Yay! We figured out how to write the original equation as a squared binomial on the left,

and a number on the right: We've completed the square. And it's easy to multiply this

out and check our work to see how we can get what we started with.

Now, to solve for x, we just take the square root of both sides, and we get:

(x + 3)2 = 25

2

4

x+ 3 = ? 5 2 2

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