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Read a fractional exponent as a root and a power, simplify radicals, and find the domain, values and translations of radical functions.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson the exponent is allowed to be a fraction. You learn to read $a^{m/n}$ as the $n$th root of $a$ raised to the power $m$, so that every exponent rule you already know applies unchanged to roots, and you use it to evaluate powers, simplify radicals and clear roots out of denominators. Then you treat a root as a function: you find the values of $x$ it accepts, evaluate it, and read off from its equation where its graph starts and which way it has been moved.
You already use the exponent rules with whole numbers: $x^3 \times x^4 = x^7$, $\frac{x^7}{x^2} = x^5$, $(x^3)^2 = x^6$, $x^{-2} = \frac{1}{x^2}$. You already know that $\sqrt{49} = 7$ and that $2^3 = 8$. Nothing new is added here. Instead the exponent is allowed to be a fraction, and every rule you already have keeps working, which is the whole point.
Radical: a root sign, $\sqrt[n]{a}$. The $n$ is the index — $2$ when it is not written — and $a$ is the radicand, the thing inside.
Rational exponent: an exponent that is a fraction, like $a^{2/3}$.
Principal root: the non-negative one. $\sqrt{9}$ means $3$, never $-3$, even though both square to $9$.
Simplest radical form: no perfect-square factor left under the root, so $\sqrt{50}$ is written $5\sqrt{2}$.
Rationalise the denominator: clear a root out of the bottom of a fraction, as $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.
A fractional exponent is a root written as a power: $$a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}.$$ So $9^{1/2} = 3$, $8^{2/3} = 2^2 = 4$, and $16^{3/4} = 2^3 = 8$. Take the root first: the numbers stay small, and $8^{2/3}$ as $\sqrt[3]{64}$ is the same answer by a harder path.
A negative exponent is a reciprocal, whether or not it is a fraction: $a^{-m/n} = \frac{1}{a^{m/n}}$, so $25^{-1/2} = \frac{1}{5}$.
Because a root is a power, every exponent rule you know applies to roots: $$x^{1/2} \times x^{1/3} = x^{5/6}, \qquad \frac{x^{3/4}}{x^{1/4}} = x^{1/2}, \qquad \left(x^{2/3}\right)^{6} = x^{4}.$$
Simplest radical form. Since $\sqrt{ab} = \sqrt{a}\sqrt{b}$, split off any perfect-square factor and let its root out: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$. To clear a root from a denominator, multiply top and bottom by it: $\frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$.
Another way: table
A two-column dictionary read in either direction. $\sqrt{x} = x^{1/2}$; $\sqrt[3]{x} = x^{1/3}$; $\sqrt[3]{x^2} = x^{2/3}$; $\frac{1}{\sqrt{x}} = x^{-1/2}$; $x\sqrt{x} = x^{3/2}$. Radicals are easier to read, powers are easier to compute with, and this table is the translation between them.
Another way: story
The fraction in the exponent is a two-part instruction: the bottom number says which root to take, the top says which power to raise it to. Doing the root first keeps every number small enough to hold in your head.
"A negative exponent gives a negative answer." It does not. $4^{-1/2} = \frac{1}{2}$, which is positive. The minus sign flips the number over; it never changes its sign.
"$\sqrt{a + b} = \sqrt{a} + \sqrt{b}$." Try $a = 9$, $b = 16$: the left side is $\sqrt{25} = 5$, the right side is $3 + 4 = 7$. Roots split over multiplication and division only, never over addition.
Reading the fraction upside down. In $a^{m/n}$ the bottom number is the root and the top number is the power. $8^{2/3}$ is $(\sqrt[3]{8})^2 = 4$; it is not $\sqrt{8^3}$ read as a cube.
The bottom number is $3$: take the cube root. $27 = 3^3$, so $\sqrt[3]{27} = 3$.
Root first, always.
The top number is $4$: raise it to the fourth power, $3^4 = 81$.
So $27^{4/3} = 81$. The other order, $\sqrt[3]{27^4} = \sqrt[3]{531441}$, gives the same $81$ from a much worse number.
Find the largest perfect square dividing $48$: it is $16$, since $48 = 16 \times 3$.
Try $4, 9, 16, 25, \ldots$ from the top down.
$\sqrt{48} = \sqrt{16}\sqrt{3} = 4\sqrt{3}$.
A root of a product is the product of the roots.
$3$ has no square factor, so nothing more comes out.
The minus flips it: $16^{-3/4} = \frac{1}{16^{3/4}}$.
The fourth root of $16$ is $2$, and $2^3 = 8$, so the value is $\frac{1}{8}$.
Evaluate $64^{2/3}$.
The value is answer.
Evaluate $36^{-1/2}$, as a fraction.
The reciprocal is answer.
Write $\sqrt{432}$ in the form $k\sqrt{12}$. What is $k$?
The coefficient is answer.
Powers of the same base are multiplied. Fill in the missing numerator.
x^(3/2) * x^(6/2) = x^(n/2)
You already read $f(x)$ as an instruction and substitute a number into it. You already solve an inequality such as $2x - 6 \ge 0$. You already know how $y = (x - 3)^2 + 1$ sits three to the right and one above $y = x^2$. Radical functions reuse all three: the only new fact is that a square root refuses to accept a negative number, and that one refusal is what makes their graphs stop.
Radical function: a function with the variable under a root, like $f(x) = \sqrt{x - 4}$.
Domain: the inputs the function accepts. Range: the outputs it produces.
Endpoint: the single point where the graph of a square-root function starts; nothing exists on the far side of it.
Translation: a slide of the whole graph, with no change of shape.
Inverse: $y = \sqrt{x}$ and $y = x^2$ for $x \ge 0$ undo each other, which is why the root graph is half a parabola lying on its side.
Domain. A square root accepts only non-negative inputs, so the domain of $f(x) = \sqrt{g(x)}$ is the solution of $g(x) \ge 0$. For $f(x) = \sqrt{3x - 12}$: $3x - 12 \ge 0$, so $x \ge 4$. An odd root, such as a cube root, has no such restriction: every real number has a real cube root.
The graph. $y = \sqrt{x}$ starts at the origin and climbs to the right, ever more slowly: it is the top half of a sideways parabola, and it is the inverse of $y = x^2$ restricted to $x \ge 0$. Its endpoint is where the whole shape is anchored, and $$y = a\sqrt{x - h} + k$$ puts that endpoint at $(h, k)$. The $h$ is inside the root, so it moves the graph sideways and against its sign: $\sqrt{x - 3}$ starts at $x = 3$. The $k$ is outside, so it moves the graph up or down, with its sign. The $a$ stretches it vertically, and flips it below the line $y = k$ when it is negative.
Equations. To solve $\sqrt{x - 1} + 2 = 5$, isolate the root, then square: $\sqrt{x - 1} = 3$, so $x - 1 = 9$ and $x = 10$. Squaring can invent solutions, so check every candidate in the original equation.
Another way: picture
The curve $y = \sqrt{x}$ drawn from the origin, rising steeply at first and then flattening. Beside it, the same curve with its endpoint dragged to $(3, 2)$ — the same shape, moved — and the region left of $x = 3$ shaded out and labelled "no real values here".
Another way: story
A square-root graph is half a parabola turned on its side. It has a starting point and only one direction to travel, which is why a radical function has a domain that stops rather than one that runs forever in both directions.
"The domain of every square root is $x \ge 0$." Only for $\sqrt{x}$ itself. For $f(x) = \sqrt{2x - 10}$ it is the inside that must be non-negative: $2x - 10 \ge 0$, so $x \ge 5$.
Shifting the wrong way. $y = \sqrt{x - 3}$ moves three to the right, not the left. Check it: the graph starts where the inside is zero, and $x - 3 = 0$ at $x = 3$.
Losing an inequality's direction. When you divide an inequality by a negative number, the sign turns around: $-2x \ge 8$ gives $x \le -4$. Dividing by a positive number leaves it alone.
What is under the root must be at or above zero: $5x + 15 \ge 0$.
This is the only condition a square root imposes.
$5x \ge -15$, and dividing by the positive $5$ keeps the direction: $x \ge -3$.
So the domain is $x \ge -3$. Note it is not $x \ge 0$: the graph starts at $(-3, 0)$.
The $+ 4$ is inside the root, so it is a sideways move against its sign: four to the left.
The inside is zero at $x = -4$.
The $- 1$ is outside, so it is a vertical move with its sign: one down.
The endpoint is at $(-4, -1)$, the domain is $x \ge -4$, and the range is $y \ge -1$.
The inside is zero when $x - 2 = 0$, that is at $x = 2$.
There the root is $0$, so $y = 3$: the endpoint is $(2, 3)$, with domain $x \ge 2$.
What is the domain of $f(x) = \sqrt{4x - 12}$?
$f(x) = 5\sqrt{x + 4} - 5$. Find $f(5)$.
The function value is answer.
The graph of $y = \sqrt{x - 6} + 4$ is the graph of $y = \sqrt{x}$ moved which way?
What is the domain of $f(x) = \sqrt{4x + 24}$?
A stone dropped down a well falls $h$ metres in $t = \sqrt{h/5}$ seconds. A stone takes $4$ seconds to reach the water. How deep is the well, in metres?
The well is answer metres deep.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Evaluate $27^{2/3}$.
The value is answer.
$f(x) = 2\sqrt{x + 2} - 6$. Find $f(2)$.
The function value is answer.
You can move between radicals and rational exponents and analyse a radical function. Evaluate $32^{2/5}$, write $\sqrt{75}$ in simplest radical form, and give the domain and starting point of $f(x) = \sqrt{2x - 8} + 1$.
8. Your turn: evaluate $16^{-3/4}$, step 2
19. Your turn: where does the graph of $y = \sqrt{x - 2} + 3$ start?, step 2