March 2018 – Manroop's Blog

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.

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