Grade 11 – Page 2 – Manroop's Blog

Week 11 – Solving Quadratic Inequalities

manroopt2015 on April 29, 2018April 29, 2018

This week in Pre-Calculus 11 we started the Solving Quadratic Inequalities unit. This week we learned about how to solve inequalities with one variable, graphing linear inequalities, and graphing quadratic inequalities.

How to Solve an Inequality with One Variable: To solve an inequality you have to isolate the variable in order to figure out the possible values of the variable, then you should test a number accordingly. When solving an inequality there are some rules that you need to remember. In order for the inequality to be true when you are diving by a negative number or multiplying by a negative number you MUST flip the inequality symbol.

NOT FLIPPING THE SIGN

Ex. $\frac{1}{-2}x>2$

$x>(2)(-2)$

$x>-4$

TEST: $x=0$

$\frac{1}{-2}\times0>2$

$0>2$

FLIPPING THE SIGN

Ex. $\frac{1}{-2}x>2$

$x<2\times-2$

$x\prec-4$

TEST: $x=-5$

$\frac{1}{-2}\times\frac{-5}{1}>2$

$\frac{-5}{-2}>2$

$2.5>2$

In the example above I showed how the inequality is not true unless you flip the sign when multiplying and dividing by a negative number. When I didn’t flip the sign, the inequality for $x$ did not work which meant that it was false. The graph in the example represents the inequality. The shaded area represents all of the possible values for x in order to make the inequality true. The shaded area and the unshaded area is separated by a broken line this represents the < sign.

How to Solve a Quadratic Inequality: To solve a quadratic inequality you have to factor the inequality and find the values of x. After finding the possible values of x you can test points between the values of x using a number line.

Ex. $x^2+12x+20>0$

$(x+10)(x+2)$

$x+10=0$ $x+2=0$

x=-10 x=-2

$-10>x>-2$

In the example above, I used a number line to solve the quadratic inequality. The first step I did was to find the zero’s of the quadratic function by factoring. Then I placed the zero’s on a number line and tested numbers between the zero’s. After testing the values using the original inequality I found out what numbers satisfied the inequality.

Week 10 – Pre Calc 11

manroopt2015 on April 21, 2018April 21, 2018

This week in Pre-Calculus 11 we prepared for the midterm. Although we didn’t learn anything new, we did review many things from the previous units that I forgot about.

Sequences and Series

Sequences and Series was the very first unit of the year. One of the concepts that I found confusing was dealing with percentages in word problems.

Ex. Billy makes $1240 a month. His boss recently told him that he will receive a 3.02% raise at the end of every month. How much will he make at the end of the first month? How much in total will he make at the end of the ninth month?

First Month:

a= $1240

$r=1.00+0.0302=1.0302$

$1240\times1.0302$

$1277.448$

$1277.45

Ninth Month:

a= $1277.45

$r=1.00+0.0302=1.0302$

n= 9

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{1277.45(1-1.0302^9}{1-1.0302})$

$S_n=\frac{1277.45(1-1.307055142)}{-.0302})$

$S_n=\frac{1277.45(-.307055142)}{-.0302})$

$S_n={-392.2475909}\div{-.0302}$

$S_n=12988.33082$

$12,988.33

In the example, the first thing I did was figure out what the common ratio was. I knew that the common ratio was above 1.0 because in the example it says that Billy will receive a 3.02% raise in his cheque, which means that his pay is increasing. Then, I found the amount he made at the end of the first month by multiplying $1240\times1.0302$ because that would be the “a” value in the second part of the question.

Infinite Sum: An infinite sum can be calculated when a geometric series converges because the series decreases a sum can be calculated. The sum can only be calculated if $-1< r< 1$ . This means that r is a fractional value and cannot be an improper fraction. The formula to calculate the infinite sum is $S_{\infty}=\frac{1}{1-r}$

Ex. 3, 9, 27, 81…

$r=\frac{t_n}{t_{n-1}}$

$r=\frac{9}{3}$

$r=3$

INFINITE SUM CANNOT BE CALCULATED

The infinite sum for the example above cannot be calculated because the common ratio is 3.

Ex. 10, -5, $\frac{5}{2}$ …

$r=\frac{t_n}{t_{n-1}}$

$r=\frac{-5}{10}$

$r=\frac{-1}{2}$

$S_{\infty}=\frac{a}{1-r}$

$S_{\infty}=\frac{10}{1-\frac{-1}{2}}$

$S_{\infty}=\frac{10}{1+\frac{1}{2}}$

$S_{\infty}=\frac{10}{\frac{3}{2}}$

$S_{\infty}=\frac{10}{1}\times\frac{2}{3}$

$S_{\infty}=\frac{20}{3}$

Absolute Value and Radicals

The absolute value and radicals unit was the second unit of the year. In this unit we learned about how the absolute value of a number is really the number of spaces it is away from 0.

Ex. $\mid-7x+x^2-10\mid$ , $x=6$

$\mid-7(6)+6^2-10\mid$

$\mid-42+36-10\mid$

$\mid-16\mid$

In the example above, I was given a value for the variable so I just had to plug in the value to find the absolute value of the expression.

Analyzing Quadratic Functions

This was the most latest unit we finished. In this unit I struggled with some of the word problem questions involving substitution.

Ex. A graph of a quadratic equation passes through (6,2) and (0,9), the axis of symmetry is x= 7. Determine an equation of the function.

STEP 1: $y=a(x-x_1)^2+q$

$y=a(x-7)^2+q$

$2=a(6-7)^2+q$

$2=a(-1)^2+q$

$2=a+q$

STEP 2: $y=a(x-x_1)^2+q$

$y=a(x-7)^2+q$

$9=a(0-7)^2+q$

$9=49a+q$

$9-49a=q$

STEP 3: $2=a+q$

$2=a+9-49a$

$2=-48a+9$

$2-9=-48a$

$-7=-48a$

$\frac{7}{48}=a$

STEP 4: $q=9-49a$

$q=9-49(\frac{7}{48})$

$q=9-\frac{343}{48}$

$q=\frac{89}{48}$

STEP 5: $y=a(x-x_1)^2+q$

$y=\frac{7}{48}(x-7)^2+\frac{89}{48}$

In the first step, I began by using standard form because I had the axis of symmetry. Then, I plugged in the point (6,2) into the x and y, this created an equation with two unknown variables. In step two, I figured out the value of “q” by using the other point given in the question (0, 9). In step three, I plugged in the value I got for “q” into the first equation I developed to find a value for “a” because we found a value for “q” we just needed to isolate “a” to find the numerical value of it. In step 4, I took the value I got for “a” and plugged it into the equation $q=9-49a$ to find the numerical value of q. Then, in step 5 I took all the information we figured out in the 4 steps and made it into an equation.

Week 9 – Pre Calc 11

manroopt2015 on April 15, 2018April 15, 2018

This week in Pre-Calculus 11 we finished the Analyzing Quadratic Functions unit. This week we learned about factored form and modelling problems with quadratic functions. Factored form is very helpful when looking for the x-intercept and it can also help with finding the axis of symmetry.

What is Factored Form? The formula for factored form is $y=a(x-x_1)(x-x_2)$ . This formula is very useful when looking for the roots of a quadratic function. You can easily convert from general form to factored form by factoring and finding the solutions of the function.

Ex. $y=x^2+6x-16$

-1 x 16= -16 -1+16= 15

-2 x 8= -16 -2+8= 6

-4 x 4= -16 -4+4= 0

$y=(x+8)(x-2)$

$x+8=0$

x= -8

$x-2=0$

x= 2

x-intercepts= (-8,0) and (2,0)

y-intercept= (0, -16)

Opens up (minimum)

Congruent to $y=x^2$

How to Find the Axis Of Symmetry: To find the axis of symmetry you have to find the average of the x-intercepts by adding them together and then dividing the sum by two. After finding the axis of symmetry you can find the vertex by plugging the x value into the equation.

Ex. $y=2x^2+7x-4$

$y=(2x-1)(x+4)$

$2x-1=0$

$2x=1$

$x=\frac{1}{2}$

$x+4=0$

$x=-4$

x-intercepts= $(-4,0)$ and $(\frac{1}{2},0)$

AOS: $x=\frac{\frac{1}{2}+-4}{2}$

AOS: $x=\frac{\frac{1}{2}+\frac{-8}{2}}{2}$

AOS: $x=\frac{\frac{-7}{2}}{2}$

AOS: $x=\frac{-7}{4}$

$y=(2(\frac{-7}{4}-1)(\frac{-7}{4}+4)$

$y=(\frac{-7}{2}-\frac{2}{2})(\frac{-7}{4}+\frac{16}{4})$

$y=(\frac{-9}{2})(\frac{9}{4}$

$y=\frac{-81}{8}$

Vertex: $(\frac{-7}{4},\frac{-81}{8})$

How to Deal with Word Problems: To deal with word problems involving quadratic functions the first step is to find a relationship between the variables. After finding the relationship between the variables find the x-intercept. Then find the axis of symmetry and plug it into the equation to find the y value. The answer of the word problem is going to involve the vertex of the quadratic function.

Ex. Find two integers with the difference of 11 and the greatest product.

$x-y=11$

$x-y-11=0$

$x-11=y$

Variable #1= x

Variable #2= x-11

Product= $(x)(x-11)$

x-intercept = (0,0)

x-11=0

x-intercept = (11,0)

AOS: $\frac{11+0}{2}$

AOS: $\frac{11}{2}$

$y=(x)(x-11)$

$y=(\frac{11}{2})(\frac{11}{2}-11)$

$y=(\frac{11}{2})(\frac{11}{2}-\frac{22}{2})$

$y=(\frac{11}{2})(\frac{-11}{2})$
$y=\frac{-121}{4}$

$(\frac{11}{2},\frac{-121}{4}$

The Impact Media Has On Mass Shootings

manroopt2015 on April 13, 2018April 13, 2018

Mass Shootings in America Are Spreading Like A Disease

This article is about the influence media has on mass shootings. Initially, this article attracted to me because of the recent mass shooting at Douglas High School in Parkland, Florida which has caused many protests and walkouts for stricter gun control. The possible motives of a mass shooter could be the popularity he/she receives in news outlets. Mass shooters are often reported about a number of times and receive “public glorification”. Within minutes the identity of the mass shooter is viral on Facebook, Instagram and Twitter. Although this problem could be solved by distributing fewer information about the shooter, but the listeners and readers would still want to be informed about the small details about the tragedy. The author mentions that a small amount of money has been spent on researching mass shootings. More research about the problem could explain the frequent recurrence of mass shootings and help prevent future shootings. The author was very insightful in his explanation and he states factual evidence of media’s role in mass shootings. It disgusts me to think that one is motivated to kill innocent lives just for recognition on television when instead you could do an act of kindness and receive respect and gratification. Although the article is mainly about media’s influence on mass shootings, at the end of the article the author also mentions the accessibility Americans have to guns. I believe that gun control is an evident reason of the repetition in mass shootings in America. In order to stop further mass shootings from occurring stricter gun control laws need to be enforced.

Week 8 – Analyzing Quadratic Functions

manroopt2015 on April 8, 2018

This week in Pre-Calculus 11 we began the Analyzing Quadratic Functions unit. This week we learned about the different properties of quadratic functions. To graph a quadratic function it is very important to have the ability to identify the properties of a quadratic function by looking at the equation.

What is General Form? General form is $ax^2+bx+c$ where c represents the y-intercept. In general form “a” represents the direction the parabola will open (if “a” is positive the parabola will open up and if “a” is negative the parabola will open down). “A” also represents the scale of the parabola and whether the parabola will be compressed or stretched.

The Scale of a Parabola: The scale of a parabola can be determined by the coefficient in front of the $x^2$ . If the coefficient is 1 the scale of the parabola will be 1-3-5 which means it is congruent to the parent function ( $y=x^2$ ). The 1-3-5 pattern means that when you start at the vertex you go one to the right and rise one and then you go one to the right and rise 3… If the coefficient of $x^2$ is not 1 then you just have to multiply 1-3-5 by the coefficient to figure out the scale.

Vertex: The vertex is the highest or lowest point of the parabola.

Axis of Symmetry: The axis of symmetry intersects the parabola at the vertex. The axis of symmetry splits the parabola exactly in half.

Minimum and Maximum Point: The minimum and maximum point of the parabola is also determined by the coefficient on $x^2$ . The minimum and maximum point is whether the vertex of the parabola is at the highest or lowest point it could possibly be at. If the vertex is at it’s minimum point that means it opens up and the coefficient on $x^2$ is positive. If the vertex is at it’s maximum point that means it opens down and the coefficient on $x^2$ is negative.

Domain: The domain of a quadratic function is all the possible values of x, the domain is alway $x\in\Re$ .

Range: The range is all the possible y values, this is dependent on the vertex.

Ex. $y=x+4x-3$

y intercept= -3

Congruent to $y=x^2$

Scale: 1-3-5

Minimum Point (opens up)

Domain: $x\in\Re$

Ex. $y=-7x^2+5x-8$

y-intercept= -8

Not congruent to $y=x^2$

Scale: -7- -21 – -35

Maximum Point (opens down)

Domain: $x\in\Re$

What is Standard Form? Standard Form is $y=a(x-p)^2+q$ . Standard form is also known as vertex form because p and q represent the vertex’s coordinates. P also tells us the horizontal translation and q also represents the vertical translation.

Ex. $y=4(x-3)^2+1$

Vertex: (3,1)

Minimum Point (opens up)

Axis of Symmetry: 3

Domain: $x\in\Re$

Range: $y\geq1$

Not Congruent to $y=x^2$

Scale: 4-12-20

Horizontal Translation: 3 to the right

Vertical Translation: 1 up

How to Convert General Form to Standard Form: You can easily go from general form to standard form by completing the square.

Ex. $y=x^2-10x+3$

$\frac{-10}{1}\times\frac{1}{2}$

$\frac{-10}{2}=-5$

$-5^2=25$

$y=x^2-10x+25-25+3$

$y=(x-5)^2-22$

Week 7 – Pre Calc 11

manroopt2015 on March 19, 2018April 14, 2018

This week in Pre-Calculus 11 we finished the solving quadratic equations unit. This week we learned about the discriminant of a quadratic equation. The discriminant can be very helpful when solving a quadratic equation because it helps determine how many solutions the equation has.

What is the Discriminant? The discriminant is from the quadratic formula. The discriminant can be calculated by using $b^2-4ac$ . Depending on whether the discriminant of a quadratic equation is positive, negative or zero you can determine how many solutions the equation may have. If the discriminant is a positive number the equation will have two solutions, if the discriminant is negative there will be no solutions, and if the discriminant is equal to zero then the equation will have one solution.

Ex. $4x^2+3x-15=0$

a= 4

b= 3

c= -15

$b^2-4ac$

$3^2-4(4)(-15)$

$9+256$

265

The example above will have 2 solutions because the discriminant is positive.

Ex. $x^2+5x+7=0$

a= 1

b= 5

c= 7

$b^2-4ac$

$5^2-4(1)(7)$

$25-28$

-3

The example above will have 0 solutions because the discriminant is negative. The discriminant is the radicand of the quadratic formula and because there cannot be negatives in the radicand there will be no solution.

Ex. $9x^2+6x+1=0$

a= 9

b= 6

c= 1

$b^2-4ac$

$6^2-4(9)(1)$

$36-36$

The example above will have one solution because the discriminant is equal to zero. Before solving for the discriminant you can simplify the equation by taking out a common factor from all of the terms.

$2x^2+6x-8$

$2(x^2+3x-4)$

a= 1

b= 3

c= -4

$b^2-4ac$

$3^2-4(1)(-4)$

$9+16$

The discriminant above is positive which means that there will be two solutions to the equation.

Week 6 – Pre Calc 11

manroopt2015 on March 11, 2018March 17, 2018

This week in Pre-Calculus 11 we continued the quadratics unit. This week we learned about different ways to solve quadratic equations and how to factor perfect square trinomials.

What is a Perfect Square Trinomial? A perfect square trinomial is when the first term and the third term are perfect squares. A perfect square trinomial cannot have a third term that is negative.

Ex. $4x^2+20x+25$

$(2x+5)^2$

The example above is a perfect square trinomial because both the first and third term are perfect squares. To factor the perfect square trinomial I found the square root of the first term and the square root of the third term.

Ex. $x^2-9x-36$

$(x-12)(x+3)$

Although the first and third term are perfect squares, the expression is NOT considered a perfect square trinomial because the last term is negative. The expression is not a perfect square trinomial but it can still be factored.

What is Completing the Square? Completing the square is when the second or third term in a perfect square trinomial is missing. When the second term is missing you have to take the third term find the square root of it and then multiply the square root by 2.

Ex. $x^2$ +___+100

$\sqrt100=10$

$10*2=20$

$x^2+20x+100$

$(x+10)^2$

When the third term of a perfect square trinomial is missing you can take the second term divide it by 2 and then square it to find the third term.

Ex. $x^2+7x$ + ___

$\frac{7}{1}\times\frac{1}{2}=\frac{7}{2}$

$(\frac{7}{2})^2=\frac{49}{4}$

$x^2+7x+\frac{49}{4}$

$(x+\frac{7}{2})^2$

What is Solving a Quadratic Equation? Solving a quadratic equation is when you have to find a value of x. When solving a quadratic formula there will always be an equals sign at the end of the equation. After factoring the equation you have to isolate the variable in order to find the value of x.

Ex. $x^2-6x-27=0$

$(x-9)(x+3)=0$

$x-9=0$

$x=9$

$x+3=0$

$x=-3$

$x=9,-3$

Solving a Quadratic Equation By Completing the Square: Another way to solve quadratic equations is by completing the square. Although this strategy can be time consuming it’s easy and it works.

Ex. $x^2+6x+23=0$

$3+3=6$

$3^2=9$

$x^2+6x+9-9+23=0$

$(x+3)^2-9+23=0$

$(x+3)^2+14=0$

$(x+3)^2=-14$

$\sqrt{(x+3)^2}=\pm\sqrt{14}$

$x+3=\pm\sqrt{14}$

$x=-3\pm\sqrt{14}$

In the example above in front of $\sqrt{14}$ is the plus or minus sign ( $\pm)$ . The $\pm$ is there because the square root of 14 can be -7 or 7. I did not calculate the value of $\sqrt{14}$ because by leaving it in the square root makes it an exact value.

The Quadratic Formula: The quadratic formula is another method to solve a quadratic equation. The formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . By plugging in the values of a, b and c, you can simply calculate the value of x.

$ax^2+bx+c=0$

Ex. $4x^2+5x=8$

$4x^2+5x-8=0$

a= 4 b= 5 c= -8

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

$x=\frac{-5\pm\sqrt{5^2-4(4)(-8)}}{2(4)}$

$x=\frac{-5\pm\sqrt{25+128}}{8}$

$x=\frac{-5\pm\sqrt{153}}{8}$

$x=\frac{-5\pm\sqrt{9*17}}{8}$

$x=\frac{-5\pm3\sqrt{17}}{8}$

How Much Technology Is Enough?

manroopt2015 on March 7, 2018April 13, 2018

Have Smartphones Destroyed a Generation?

This article is about the repercussions of technology in millennials lives because of the excessive use of technology. I was interested in this article because I believe that my generation spends too much time on technology. By spending an unreasonable amount of time on technology we have missed matchless experiences and opportunities. Today, people are glued to their phones and often avoid face-to-face interactions. The author compared and contrasted different generations insightfully by using data from previous generations. I enjoyed reading about the authors personal opinion about iGen and her childhood, “Unlike the teens of my generation, who might have spent an evening tying up the family landline with gossip”. The article reveals millennials dependence on technology which can be seen universally. For example, when you feel your phone vibrating, you have an irresistible urge to check the notification you received. Although technology has brought the world innovation and ease, it has also taken away many aspects of life.

Week 5 – Solving Quadratic Equations

manroopt2015 on March 4, 2018March 10, 2018

In the fifth week of Pre-Calculus 11 we began the Quadratics unit. This week we reviewed factoring polynomials and learned a few new strategies about factoring.

What is Factoring? Factoring is taking an expression and breaking it up into pieces so that you can multiply those pieces together to get the original number. For example, the number 9 can be factored to $9*1$ or $3*3$ . To factor polynomials it’s helpful to think about the acronym CDPEU.

C- Is there anything COMMON within the terms?

Ex. $7x^3+14x$

$7x(x^2+2)$

In the expression above, the greatest common factor is $7x$ because $7x$ can factor into both of the terms.

D- Is there a DIFFERENCE of squares? A difference of squares requires the expression to be a binomial, the terms have to be subtracting (difference) and both of the terms have to be perfect squares.

Ex. $100x^2-4$

$(10x+2)(10x-2)$

P- Does the expression have the right PATTERN? To factor a trinomial the expression has to have the correct pattern by having $x^2$ , x, and a number.

Ex. $3+y-x$

The expression above cannot be factored because it does not have the pattern $x^2$ , x, and a number. It also does not have any common terms.

Ex. $x^2+7x+12$

$1*12=12$ $1+12=13$

$2*6=12$ $2+6=8$

$3*4=12$ $3+4=7$

$(x+4)(x+3)$

The trinomial above can be simplified because it has the pattern we are looking for. To find out what numbers I needed to use I listed all of the factors of 12. After listing the factors of 12 I figured out which one of the factors adds up to the middle term (7).

E– Is the expression EASY to factor? To determine the complexity of an expression you have to check if the $x^2$ has a coefficient. If the $x^2$ does not have a coefficient then it is an easy expression to factor.

Ex. $x^2-12x+20$

$-1*-20=20$ $-1+-20=-21$

$-2*-10=20$ $-2+-10=-12$

$-4*-5=20$ $-4+-5=-9$

$(x-10)(x-2)$

U- Is the expression UGLY to factor? A trinomial that’s difficult to factor is when the $x^2$ has a coefficient.

Ex. $2x^2-9x+4$

$(2x-1)(x-4)$

To factor the expression above I multiplied the first and the last term together. After multiplying the two terms together I got $8x^2$ , then I began to list all the factors of $8x^2$ . After figuring out the factors of $8x^2$ I chose the two factors that add up to -9x. Lastly, I placed the two factors in the chart above and found the GCF of the terms that were horizontal and vertical from each other.

Week 4 – Pre Calc 11

manroopt2015 on February 24, 2018February 24, 2018

In the fourth week of Pre-Calculus 11 we continued the absolute value and radicals unit. This week we learned about addition and subtraction of radicals.

Addition and Subtraction of Radicals: To add and subtract radicals, the radicands have to be of the same value. Not only do the radicands have to be the same, but the index also has to be the same. The coefficients in front of the radical are added or subtracted, not the radicand.

Ex. $\sqrt{23}+\sqrt[4]{23}$

The example above cannot be added together because the index’s are not the same. Although, the radicals have the same radicand, the index of the radicals are different.

Ex. $2\sqrt{21}+\sqrt{21}$

$3\sqrt{21}$

The expression above could be added together because the radicand and the index are the same. I added the coefficients of the radical together and not the value inside the radical.

Ex. $\sqrt{20}+8\sqrt{18}-2\sqrt{50}+\sqrt{45}$

$\sqrt{4\times5}+8\sqrt{9\times2}-2\sqrt{25\times2}+\sqrt{9\times5}$

$2\sqrt{5}+24\sqrt{2}-10\sqrt{2}+3\sqrt{5}$

$5\sqrt{5}+14\sqrt{2}$

In the example above, before adding/subtracting the radicals I simplified everything inside the radical. By simplifying the radical first, I was able to tell whether I could add/subtract the radicals together.

How to Multiply Radicals: To multiply radicals you have to multiply the coefficients by the coefficients and multiply the radicand by the radicand.

Ex. $3\sqrt{5}\times7\sqrt{2}$

$21\sqrt{10}$

It’s best to simplify the radical before multiplying, if possible. By simplifying the radical beforehand, it’ll be easier to multiply the radicals together because the numbers will be smaller.

Ex. $2\sqrt{5}(\sqrt{75}-7\sqrt{8})$

$2\sqrt{5}(\sqrt{25\times3}-7\sqrt{4\times2})$

$2\sqrt{5}(5\sqrt{3}-14\sqrt{2})$

$10\sqrt{15}-28\sqrt{10}$

In the example above, I simplified the radicals to make the numbers more manageable to multiply. After simplifying the radicals I began to distribute inside the brackets.

How to Divide Radicals: When dividing a question that involves radicals you can only divide radicals by radicals and coefficients by coefficients.

Ex. $\frac{\sqrt{2}}{4}$

The example above cannot be simplified any further. Although 4 and 2 have a common factor of 2, they cannot be simplified because they are not in the same form.

Ex. $\frac{6\sqrt{10}}{9\sqrt{5}}$

$\frac{2\sqrt{2}}{3\sqrt{1}}$

$\frac{2\sqrt{2}}{3}$

The most important rule for dividing radicals is that a radical cannot be left in the denominator of a fraction. If the denominator does contain a radical you have to rationalize the denominator.

Rationalize the Denominator: To rationalize the denominator with a radical you have to multiply the whole fraction by the denominator.

Ex. $\frac{\sqrt{35}}{\sqrt{10}}$

$\frac{\sqrt{7}}{\sqrt{2}}$

$\frac{\sqrt{7}}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}$

$\frac{\sqrt{14}}{\sqrt{4}}$

$\frac{\sqrt{14}}{2}$

When the denominator is a binomial containing a radical you have to multiply by the conjugate. To find the conjugate of a binomial you have to flip the sign in the middle of the binomial. The conjugate of $6+\sqrt{3}$ is $6-\sqrt{3}$ .