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1. Rational exponents in exponential functions

Warm-up

Three quick questions from earlier lessons.

Discover

A bacteria population doubles every day. By what factor does it grow in half a day?

Hint: whatever the factor $f$ is, growing for two half-days must double the population, so $f \cdot f = 2$ .

Key idea

Exponents that are fractions are roots.

b^{\frac{1}{n}} = \sqrt[n]{b}

and

b^{\frac{m}{n}} = \left(\sqrt[n]{b}\right)^m

So the half-day factor is $2^{\frac12} = \sqrt{2} \approx 1.41$ . A negative exponent means a reciprocal: $b^{-r} = \dfrac{1}{b^{r}}$ .

Worked example

Evaluate $8^{\frac{2}{3}}$ .

8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2

The denominator 3 means a cube root; the numerator 2 means a square.

= 2^2

The cube root of 8 is 2.

= 4

Square it.

Your turn, guided

Your turn: evaluate $27^{\frac{2}{3}}$ . Fill in the missing line.

27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2

Cube root, then square.

= 9

3² = 9.

Practice ★

Evaluate $36^{ \frac{ 5 }{2} }$ .
Evaluate $81^{ \frac{ 3 }{2} }$ .
Evaluate $81^{ \frac{ 5 }{2} }$ .
Evaluate $4^{ \frac{ 3 }{2} }$ .
Evaluate $49^{ \frac{ 5 }{2} }$ .

Watch out!

Practice ★★

Evaluate $125^{ -\frac{ 2 }{3} }$ .
Evaluate $125^{ -\frac{ 1 }{3} }$ .
Evaluate $64^{ -\frac{ 1 }{3} }$ .
Evaluate $8^{ -\frac{ 1 }{3} }$ .
Evaluate $27^{ -\frac{ 1 }{3} }$ .

Exit ticket

Three questions to check you have got it.

Evaluate $125^{ -\frac{ 2 }{3} }$ .
Evaluate $125^{ -\frac{ 1 }{3} }$ .
Evaluate $64^{ -\frac{ 1 }{3} }$ .

Summary card

Rational exponents

$b^{\frac{1}{n}} = \sqrt[n]{b}$ ; $b^{\frac{m}{n}} = \left(\sqrt[n]{b}\right)^m$ .
$b^{-r} = \dfrac{1}{b^{r}}$ : the value stays positive.
A factor $b$ per unit means a factor $b^{\frac12}$ per half-unit.

2. Solving radical equations; extraneous solutions

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $\sqrt{x} = -3$ . Squaring gives $x = 9$ . Does 9 work?

No, $\sqrt{9} = 3$
Yes

Key idea

Undo the root, then check.

To solve a radical equation, isolate the radical, raise both sides to the matching power (square for

\sqrt{\ }

, cube for

\sqrt[3]{\ }

), and solve. Squaring can create extraneous solutions: always check in the original equation. Cubing does not.

Worked example

$\sqrt{x + 6} = x$ .

x + 6 = x^2

Square.

x^2 - x - 6 = 0

, so

x = 3

x = -2

Solve.

Only

x = 3

√9 = 3 ✓; √4 = 2 ≠ −2 ✗.

Your turn, guided

Your turn: $\sqrt{x - 1} = 4$ .

x = 17

x − 1 = 16.

Practice ★

Solve $\sqrt{x + 2} = 8$ .
Solve $\sqrt{x + 8} = 7$ .

Solve $\sqrt[3]{x - 8} = 3$ .

Watch out!

Practice ★★

Solve $\sqrt{x + 6} = x$ . (Squaring gives two candidates; enter the one that works.)
Solve $\sqrt{x + 30} = x$ . (Squaring gives two candidates; enter the one that works.)

Exit ticket

Three questions to check you have got it.

Solve $\sqrt{x + 2} = 8$ .
Solve $\sqrt{x + 8} = 7$ .
Solve $\sqrt{x + 4} = 6$ .

Summary card

Radical equations

Isolate → square (or cube) → solve → check.

Watch out: squaring can add extraneous solutions.

3. The number i and complex numbers

Warm-up

Three quick questions from earlier lessons.

Discover

No real number squares to $-1$ . Mathematicians define $i$ with $i^2 = -1$ . What is $(3i)^2$ ?

$-9$
$9$

Key idea

A new number that makes negative square roots possible.

i^2 = -1

. For

a > 0

\sqrt{-a} = i\sqrt{a}

. A complex number is

a + bi

with real part

a

and imaginary part

b

. Powers of

i

cycle:

i, -1, -i, 1, \dots

(period 4).

Worked example

Simplify $\sqrt{-49}$ and $i^{10}$ .

\sqrt{-49} = 7i

√49 · √−1.

i^{10} = i^2

10 = 4·2 + 2.

i^{10} = -1

i² = −1.

Your turn, guided

Your turn: $i^7$ .

-i

7 = 4 + 3: i³ = −i.

Practice ★

Write $\sqrt{-121}$ in terms of $i$ .
Write $\sqrt{-4}$ in terms of $i$ .

Which number is real?
1. $(9i)^2$
2. $\sqrt{-9}$
3. $9 + i$

Watch out!

Practice ★★

Simplify $i^{ 10 }$ .
Simplify $i^{ 13 }$ .
Simplify $i^{ 5 }$ .

Exit ticket

Three questions to check you have got it.

Write $\sqrt{-121}$ in terms of $i$ .
Write $\sqrt{-4}$ in terms of $i$ .
Write $\sqrt{-25}$ in terms of $i$ .

Summary card

The number i

$i^2 = -1$ ; $\sqrt{-a} = i\sqrt{a}$ ; powers of $i$ repeat every 4.

Watch out: $\sqrt{-16}$ is $4i$ , not $-4$ .

4. Adding, subtracting and multiplying complex numbers

Warm-up

Three quick questions from earlier lessons.

Discover

$(2 + 3i) + (4 + 5i)$ . Treat $i$ like a variable. What do you get?

$6 + 8i$
$14$

Key idea

Calculate as with polynomials in $i$ , then use $i^2 = -1$ .

(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i

(a + bi)(c + di) = (ac - bd) + (ad + bc)i

, since

bd\,i^2 = -bd

. The conjugate of

a + bi

a - bi

, and

(a + bi)(a - bi) = a^2 + b^2

Worked example

$(2 + 3i)(1 - 4i)$ .

2 - 8i + 3i - 12i^2

Distribute.

2 - 5i + 12

i² = −1.

14 - 5i

Combine.

Your turn, guided

Your turn: $(3 + 2i)(3 - 2i)$ .

13

9 + 4.

Practice ★

Add: $(9 + 8i) + (2 + 1i)$ .
Add: $(3 + 4i) + (9 + 2i)$ .

Multiply: $(2 + 8i)(2 - 8i)$ .
Multiply: $(7 + 2i)(7 - 2i)$ .

Watch out!

Practice ★★

Expand $(3 + 6i)^2$ .
Expand $(1 + 6i)^2$ .

Exit ticket

Three questions to check you have got it.

Multiply: $(2 + 8i)(2 - 8i)$ .
Multiply: $(7 + 2i)(7 - 2i)$ .
Multiply: $(8 + 4i)(8 - 4i)$ .

Summary card

Complex arithmetic

Add parts separately; multiply out and use $i^2 = -1$ ; conjugates give a real product.

Watch out: $i^2$ is $-1$ , not $1$ .

5. Quadratic equations with complex solutions

Warm-up

Three quick questions from earlier lessons.

Discover

$x^2 + 9 = 0$ has no real solution. With $i$ , which numbers solve it?

$x = 3i$ and $x = -3i$
$x = 3$ and $x = -3$

Key idea

Every quadratic has solutions among the complex numbers.

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

. If the discriminant

b^2 - 4ac < 0

, the solutions are two non-real complex conjugates

p \pm qi

; if

= 0

, one real solution; if

> 0

, two real solutions.

Worked example

$x^2 - 4x + 13 = 0$ .

Discriminant

= -36

16 − 52.

x = \dfrac{4 \pm 6i}{2}

√(−36) = 6i.

x = 2 \pm 3i

Divide both parts by 2.

Your turn, guided

Your turn: the discriminant of $x^2 + 2x + 5 = 0$ .

-16

: two non-real solutions

4 − 20.

Practice ★

$x^2 + 36 = 0$ . Enter the solution with positive imaginary part.
$x^2 + 81 = 0$ . Enter the solution with positive imaginary part.

Find the discriminant of $1x^2 + 7x + 4 = 0$ .

Watch out!

Practice ★★

$x^2 - 4x + 20 = 0$ . Enter the solution with positive imaginary part.
$x^2 - 6x + 34 = 0$ . Enter the solution with positive imaginary part.

Boss problem. Solve $x^2 - 10x + 29 = 0$ with the quadratic formula.

Exit ticket

Three questions to check you have got it.

$x^2 + 36 = 0$ . Enter the solution with positive imaginary part.
$x^2 + 81 = 0$ . Enter the solution with positive imaginary part.
$x^2 + 25 = 0$ . Enter the solution with positive imaginary part.

Summary card

Complex solutions

Discriminant < 0 → $x = p \pm qi$ (conjugates).

Watch out: divide the whole numerator by $2a$ .

6. Complex zeros and conjugate pairs

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $x^2 = -9$ . Is there a real number whose square is $-9$ ?

No real number; the solutions are $3i$ and $-3i$
Yes: $3$ and $-3$

Key idea

The number $i$ has $i^2 = -1$ . Complex numbers $a + bi$ give every quadratic its zeros.

For

ax^2 + bx + c

: if

b^2 - 4ac < 0

, there are no real zeros but two complex zeros

\dfrac{-b \pm i\sqrt{4ac - b^2}}{2a}

. A polynomial with real coefficients has complex zeros in conjugate pairs:

p + qi

and

p - qi

Worked example

Solve $x^2 - 4x + 13 = 0$ .

b^2 - 4ac = 16 - 52 = -36

The discriminant decides.

x = \dfrac{4 \pm 6i}{2}

√(−36) = 6i.

x = 2 + 3i

x = 2 - 3i

Divide both parts by 2.

Your turn, guided

Your turn: $x^2 - 2x + 5 = 0$ .

b^2 - 4ac = -16

4 − 20.

x = 1 + 2i

x = 1 - 2i

√(−16) = 4i, and (2 + 4i) ÷ 2 = 1 + 2i. Type this first solution.

Practice ★

How many real zeros does $x^2 + 5x + 10$ have?
1. Two
2. One
3. None: two complex zeros
How many real zeros does $x^2 + 4x + 9$ have?
1. Two
2. One
3. None: two complex zeros
How many real zeros does $x^2 + 2x + 20$ have?
1. Two
2. One
3. None: two complex zeros

Watch out!

Practice ★★

Solve $x^2 - 2x + 5 = 0$ . Its solutions are $u + vi$ and $u - vi$ , with $v > 0$ .

Solve $x^2 - 8x + 25 = 0$ . Type each solution as $a + bi$ .
Solve $x^2 - 8x + 32 = 0$ . Type each solution as $a + bi$ .

A polynomial with real coefficients has the zero $4 + 4i$ . Which other number must be a zero?
1. $4 - 4i$
2. $-4 - 4i$
3. $-4 + 4i$
A polynomial with real coefficients has the zero $8 + 7i$ . Which other number must be a zero?
1. $8 - 7i$
2. $-8 - 7i$
3. $-8 + 7i$
A polynomial with real coefficients has the zero $4 + 5i$ . Which other number must be a zero?
1. $4 - 5i$
2. $-4 - 5i$
3. $-4 + 5i$

Exit ticket

Three questions to check you have got it.

Solve $x^2 - 8x + 25 = 0$ . Type each solution as $a + bi$ .
Solve $x^2 - 8x + 32 = 0$ . Type each solution as $a + bi$ .
Solve $x^2 - 4x + 5 = 0$ . Type each solution as $a + bi$ .

Summary card

Complex zeros

$i^2 = -1$ , so $\sqrt{-k^2} = ki$ .
$b^2 - 4ac < 0$ : no real zeros, two complex ones.
Real coefficients: $p + qi$ and $p - qi$ come together.

Watch out: the conjugate changes only the sign of the $i$ part.

Answers: Complex numbers and rational exponents

1. Rational exponents in exponential functions

$7776$
$729$
$59049$
$8$
$16807$

$\frac{1}{25}$
$\frac{1}{5}$
$\frac{1}{4}$
$\frac{1}{2}$
$\frac{1}{3}$

$\frac{1}{25}$
$\frac{1}{5}$
$\frac{1}{4}$

2. Solving radical equations; extraneous solutions

$62$
$41$

$35$

$62$
$41$
$32$

3. The number i and complex numbers

$11 i$
$2 i$

$(9i)^2$

$-1$
$i$
$i$

$11 i$
$2 i$
$5 i$

4. Adding, subtracting and multiplying complex numbers

$11 + 9 i$
$12 + 6 i$

$68$
$53$

$-27 + 36 i$
$-35 + 12 i$

$68$
$53$
$80$

5. Quadratic equations with complex solutions

$6 i$
$9 i$

$33$

$2 + 4 i$
$3 + 5 i$

$5 + 2 i$

$6 i$
$9 i$
$5 i$

6. Complex zeros and conjugate pairs

None: two complex zeros
None: two complex zeros
None: two complex zeros

$4 + 3 i,\ 4 - 3 i$
$4 + 4 i,\ 4 - 4 i$

$4 - 4i$
$8 - 7i$
$4 - 5i$

$4 + 3 i,\ 4 - 3 i$
$4 + 4 i,\ 4 - 4 i$
$2 + i,\ 2 - i$