Suppose that we wish to evaluate the expression \sqrt x + 3 for the value x=9. This type of expression is called a radical expression because it contains the radical symbol \sqrt{\phantom 0}. Remember that \sqrt x means "the positive square root of the number x. "

To evaluate this expression at the given value, we replace x with 9 in the given expression and simplify.

\begin{align*} \sqrt x + 3 &=\\[3pt] \sqrt 9 + 3 &=\\[3pt] \sqrt{3^2} + 3 &=\\[3pt] 3 + 3 &=\\[3pt] 6 \end{align*}

Important: You may have heard somewhere that \sqrt{9} = \pm 3, because (-3)^2=9 and (3)^2 = 9. However, when we write down \sqrt{9}, we always mean "the positive square root of 9 ".

If we want the negative square root of 9 , we write -\sqrt{9} = -3. Sometimes, we want both, in which case we would write \pm\sqrt{9}=\pm 3.

FLAG

Evaluate \sqrt{x} + 2x -1 when x=16.

EXPLANATION

We replace x with 16 in the given expression, and simplify:

\begin{align*} \sqrt{x} + 2x -1 &=\\[3pt] \sqrt{16} + 2\cdot 16 -1 &=\\[3pt] \sqrt{4^2} + 32 -1 &=\\[3pt] 4 + 32 -1 &=\\[3pt] 35 \end{align*}

FLAG

Evaluate $x(\sqrt{x}-1)$ when $x=9.$

a
$36$
b
$9$
c
$27$
d
$18$
e
$-18$

Evaluate $2\sqrt{2x} + 8$ when $x=18.$

a
$30$
b
$32$
c
$16$
d
$20$
e
$4$

Evaluate \sqrt{3x-1} for x=3.

EXPLANATION

We replace x with 3 in the given expression, and simplify:

\begin{align*} \sqrt{3x-1} &=\\[3pt] \sqrt{3\cdot 3 -1}&=\\[3pt] \sqrt{9 -1}&=\\[3pt] \sqrt{8}& \end{align*}

We can simplify even further using the product rule:

\begin{align*} \sqrt{8}&=\\[3pt] \sqrt{4\cdot 2}&=\\[3pt] \sqrt{2^2\cdot 2}&=\\[3pt] \sqrt{2^2}\cdot \sqrt 2&=\\[3pt] 2\sqrt 2 \end{align*}

FLAG

Evaluate $\sqrt{5x+4}$ for $x=8.$

a
$4\sqrt {11}$
b
$2\sqrt {11}$
c
$2\sqrt {22}$
d
$8\sqrt {11}$
e
$4\sqrt {22}$

Evaluate $\sqrt{7x-4}$ for $x=7.$

a
$3\sqrt {15}$
b
$3\sqrt 3$
c
$9\sqrt 5$
d
$3\sqrt 5$
e
$5\sqrt {3}$

Given that f(x) = \dfrac{\sqrt{9x+3}}{\sqrt{x-2}}, find f(5).

EXPLANATION

We replace x with 5 in the given expression, and simplify:

\begin{align*} f(5) & = \dfrac{\sqrt{9(5)+3}}{\sqrt{(5)-2}} \\[5pt] &= \dfrac{\sqrt{48}}{\sqrt{3}} \\[5pt] &= \sqrt{\dfrac{48}{3}}\\[5pt] &= \sqrt{16}\\[5pt] &= \sqrt{4^2}\\[5pt] &= 4 \end{align*}

FLAG

Given $f(x) = \dfrac{\sqrt{x}}{\sqrt{8-x}},$ find $f(6).$

a
$\sqrt{2}$
b
$1$
c
$2$
d
$\dfrac{1}{2}$
e
$\sqrt{3}$

Given $h(x) = \dfrac{\sqrt{x+6}}{\sqrt{x}},$ find $h(2).$

a
$6$
b
$2\sqrt{2}$
c
$2$
d
$4\sqrt{2}$
e
$8$

Evaluate \sqrt[3]{x+6} at x=2.

EXPLANATION

We replace x with 2 in the given expression, and simplify as much as possible: \begin{align*} \sqrt[3]{x+6} &=\\[3pt] \sqrt[3]{2+6} &=\\[3pt] \sqrt[3]{8} &=\\[3pt] \sqrt[3]{2^3} &=\\[3pt] 2 \end{align*}

FLAG

Evaluate $\sqrt[3]{x+3}$ at $x=5.$

a
$4$
b
$\sqrt{2}$
c
$2$
d
$1$
e
$3$

Evaluate $\sqrt[3]{7x+5}$ at $x=7.$

a
$9\sqrt[3]{2}$
b
$3\sqrt[3]{2}$
c
$18$
d
$3\sqrt[3]{6}$
e
$27\sqrt[3]{2}$
Flag Content
Did you notice an error, or do you simply believe that something could be improved? Please explain below.
SUBMIT
CANCEL