Practice questions

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Radical expressions and rational numbers as exponents.

20. Simplify the following expressions.

(a) $\sqrt[5]{s^3}\:\sqrt[4]{s}$ (Write the answer in radical notation.)

(b) $\sqrt{108}$

(d) $3x\sqrt[3]{xy^2}-2\sqrt[3]{8x^4y^2}$

(e) $\sqrt{x^2-4x+4}$; $x\ge 0$

Solution:

(a).

First write the expression in exponential notation:

$=s^{\frac{3}{5}}\:s^{\frac{1}{4}}$

$=s^{\frac{3}{5}+\frac{1}{4}}$

$=s^{\frac{17}{20}}$

Write in radical form:

$=\sqrt[20]{s^{17}}$

(b).

Factoring what is inside the radical sign:

=$\sqrt{4\cdot 9\cdot 3}$

=$2\cdot 3\sqrt{3}$

=$6\sqrt{3}$

(c).

=$-(-5)a^2b^3c^4$

=$5a^2b^3c^4$

(d).

=$3x\sqrt[3]{xy^2}-2\cdot 2 x\sqrt[3]{xy^2}$

=$3x\sqrt[3]{xy^2}-4x\sqrt[3]{xy^2}$

=$-x\sqrt[3]{xy^2}$

(e).

$\sqrt{x^2-4x+4}$;

Factor the expression inside the radical sign:

=$\sqrt{(x-2)^2}$

=$|x-2|$

Note that, I put the absolute value symbol as the radical sign represents a positive number. It is given, $x\ge 0$, that includes for example, $1$. If you substitute $x=1$, you will get a negative number if you don't put the absolute value symbol.

21. Rationalize the denominator for the following expressions.

(a) $\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}}$

(b) $\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}$

(a).

Multiply and divide the conjugate of the denominator (change the sign of the second term to get the conjugate):

$=\dfrac{(\sqrt{2}-\sqrt{3})\cdot (\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})\cdot (\sqrt{6}+\sqrt{5})}$

Do multiplication in the numerator and remove the parentheses. The denominator, write as a difference of squares:

$=\dfrac{\sqrt{2}\cdot \sqrt{6}+\sqrt{2}\cdot \sqrt{5}-\sqrt{3}\cdot \sqrt{6}-\sqrt{3}\cdot \sqrt{5}}{(\sqrt{6})^2-(\sqrt{5})^2}$

$=\dfrac{\sqrt{12}+\sqrt{10}-\sqrt{18}-\sqrt{15}}{6-5}$

Factor out the square roots if possible,

$=\dfrac{\sqrt{4\cdot 3}+\sqrt{10}-\sqrt{9\cdot 2}-\sqrt{15}}{1}$

$=2\sqrt{3}+\sqrt{10}-3\sqrt{2}-\sqrt{15}$

22. Multiply.

(a) $(3\sqrt{7}+2\sqrt{5})(2\sqrt{7}-4\sqrt{5})$

(b) $(3-\sqrt{2})^2$

Solution:

(a).

Multiply and remove the parentheses:

$=3\sqrt{7}\cdot 2\sqrt{7}-3\sqrt{7}\cdot 4\sqrt{5}+2\sqrt{5}\cdot 2\sqrt{7}-2\sqrt{5}\cdot 4\sqrt{5}$

Multiply the numbers outside the square roots and combine the square roots:

$=6\sqrt{49}-12\sqrt{35}+4\sqrt{35}-8\sqrt{25}$

$=6\cdot 7-12\sqrt{35}+4\sqrt{35}-8\cdot 5$

$=42-8\sqrt{35}-40$

$=2-8\sqrt{35}$

23. Solve the following equations.

(a) $\sqrt[3]{x-8}+3=0$

(b) $5=\sqrt{7x-3}$

Solution:

(a).

Isolate the cube root:

$\sqrt[3]{x-8}=\color{red}-3$

Take the third power on both sides to remove the radical sign:

$(\sqrt[3]{x-8})^3=(-3)^3$

$x-8=-27$

Solve for $x$:

$x=-19$

Thus, the solution is $x=-19$.

(b).

Square on both sides to remove the square root:

$5^2=(\sqrt{7x-3})^2$

$25=7x-3$

Solving for $x$:

$x=4$

Since the original equation has a square root (an even root), you need to check whether $x=4$ is the solution. Substituting $x=4$ in the original equation:

$5=\sqrt{7\cdot 4-3}$

$5=\sqrt{25}$

$5=5$

This is true. So, $x=4$ is the solution.

(c).

The equation can be written in radical form as

$\sqrt[3]{(2w-1)^2}-\sqrt[3]{w}=0$

Move the second term on the left hand side to the right hand side:

$\sqrt[3]{(2w-1)^2}=\sqrt[3]{w}$

Now we can remove the radical sign by taking the third power on both sides:

$\left(\sqrt[3]{(2w-1)^2}\right)^3=(\sqrt[3]{w})^3$

$(2w-1)^2=w$

Use the formula, $(a-b)^2=a^2-2ab+b^2$ to expand the left hand side:

$(2w)^2-2\cdot 2w\cdot 1+1^2=w$

$4w^2-4w+1=w$

Move $w$ to the left hand side and solve the quadratic equation:

$4w^2-4w+1\color{red}-w$$=0$

$4w^2-5w+1=0$

Solve by factoring:

$4w^2-4w-w+1=0$

$4w(w-1)-(w-1)=0$

$(w-1)(4w-1)=0$

By zero-product rule:

$w-1=0$ or $4w-1=0$

$w=1$ or $w=1/4$