# 1.3 Radicals and rational exponents  (Page 3/11)

 Page 3 / 11

## Adding square roots

Add $\text{\hspace{0.17em}}5\sqrt{12}+2\sqrt{3}.$

We can rewrite $\text{\hspace{0.17em}}5\sqrt{12}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}5\sqrt{4·3}.\text{\hspace{0.17em}}$ According the product rule, this becomes $\text{\hspace{0.17em}}5\sqrt{4}\sqrt{3}.\text{\hspace{0.17em}}$ The square root of $\text{\hspace{0.17em}}\sqrt{4}\text{\hspace{0.17em}}$ is 2, so the expression becomes $\text{\hspace{0.17em}}5\left(2\right)\sqrt{3},$ which is $\text{\hspace{0.17em}}10\sqrt{3}.\text{\hspace{0.17em}}$ Now we can the terms have the same radicand so we can add.

$10\sqrt{3}+2\sqrt{3}=12\sqrt{3}$

Add $\text{\hspace{0.17em}}\sqrt{5}+6\sqrt{20}.$

$13\sqrt{5}$

## Subtracting square roots

Subtract $\text{\hspace{0.17em}}20\sqrt{72{a}^{3}{b}^{4}c}\text{\hspace{0.17em}}-14\sqrt{8{a}^{3}{b}^{4}c}.$

Rewrite each term so they have equal radicands.

$\begin{array}{ccc}\hfill 20\sqrt{72{a}^{3}{b}^{4}c}& =& 20\sqrt{9}\sqrt{4}\sqrt{2}\sqrt{a}\sqrt{{a}^{2}}\sqrt{{\left({b}^{2}\right)}^{2}}\sqrt{c}\hfill \\ & =& 20\left(3\right)\left(2\right)|a|{b}^{2}\sqrt{2ac}\hfill \\ & =& 120|a|{b}^{2}\sqrt{2ac}\hfill \end{array}$
$\begin{array}{ccc}\hfill 14\sqrt{8{a}^{3}{b}^{4}c}& =& 14\sqrt{2}\sqrt{4}\sqrt{a}\sqrt{{a}^{2}}\sqrt{{\left({b}^{2}\right)}^{2}}\sqrt{c}\hfill \\ & =& 14\left(2\right)|a|{b}^{2}\sqrt{2ac}\hfill \\ & =& 28|a|{b}^{2}\sqrt{2ac}\hfill \end{array}$

Now the terms have the same radicand so we can subtract.

$120|a|{b}^{2}\sqrt{2ac}-28|a|{b}^{2}\sqrt{2ac}=92|a|{b}^{2}\sqrt{2ac}$

Subtract $\text{\hspace{0.17em}}3\sqrt{80x}\text{\hspace{0.17em}}-4\sqrt{45x}.$

$0$

## Rationalizing denominators

When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator .

We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical.

For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is $\text{\hspace{0.17em}}b\sqrt{c},$ multiply by $\text{\hspace{0.17em}}\frac{\sqrt{c}}{\sqrt{c}}.$

For a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is $\text{\hspace{0.17em}}a+b\sqrt{c},$ then the conjugate is $\text{\hspace{0.17em}}a-b\sqrt{c}.$

Given an expression with a single square root radical term in the denominator, rationalize the denominator.

1. Multiply the numerator and denominator by the radical in the denominator.
2. Simplify.

## Rationalizing a denominator containing a single term

Write $\text{\hspace{0.17em}}\frac{2\sqrt{3}}{3\sqrt{10}}\text{\hspace{0.17em}}$ in simplest form.

The radical in the denominator is $\text{\hspace{0.17em}}\sqrt{10}.\text{\hspace{0.17em}}$ So multiply the fraction by $\text{\hspace{0.17em}}\frac{\sqrt{10}}{\sqrt{10}}.\text{\hspace{0.17em}}$ Then simplify.

Write $\text{\hspace{0.17em}}\frac{12\sqrt{3}}{\sqrt{2}}\text{\hspace{0.17em}}$ in simplest form.

$6\sqrt{6}$

Given an expression with a radical term and a constant in the denominator, rationalize the denominator.

1. Find the conjugate of the denominator.
2. Multiply the numerator and denominator by the conjugate.
3. Use the distributive property.
4. Simplify.

## Rationalizing a denominator containing two terms

Write $\text{\hspace{0.17em}}\frac{4}{1+\sqrt{5}}\text{\hspace{0.17em}}$ in simplest form.

Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of $\text{\hspace{0.17em}}1+\sqrt{5}\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}1-\sqrt{5}.\text{\hspace{0.17em}}$ Then multiply the fraction by $\text{\hspace{0.17em}}\frac{1-\sqrt{5}}{1-\sqrt{5}}.$

Write $\text{\hspace{0.17em}}\frac{7}{2+\sqrt{3}}\text{\hspace{0.17em}}$ in simplest form.

$14-7\sqrt{3}$

## Using rational roots

Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.

#### Questions & Answers

write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3