Week 6 blog post

Posted on October 13, 2018 by saejino2016

On this week, I learned Solving Quadratic Equations, Using Square Roots to Solve Quadratic Equations, and Using the Quadratic Formula to Solve Quadratic Equations. Its form is $ax^2$ + bx+ c=0

Solving Quadratic Equations.The zero product property – a*b=0 (x+2)(x-7)=0 x+2=0, x-7=0 x=-2, 7 (2x+1)(3x-5)=0 2x+1=0, 3x-5=0 x= $\frac{-1}{2}$ , $\frac{5}{3}$ $x^2$ – 81=0 -> factor (x+9)(x-9)=0 x= 9, -9 $10x^2$ – 90x=0 -> Find a common factor 10x(x-9)=0 x= 0, 9 $x^2$ -9x -22 = 0 -> factor (x-11)(x+2)=0 x=11, -2

If the right side of the equation is not equal to zero, so expand the left side. Collect all terms on the left side to get 0 on the right side. Factor the trinomial. Use the zero product property.

ex) (3x+1)(x-6)=22 (3x+1)(x-6)-22=0 $3x^2$ – 7x-6-22=0 $3x^2$ -7x-28=0 -> factor (3x+4)(x-7)=0 x= $\frac{-4}{3}$ , 7

Using Square Roots to solve Quadratic Equations

Solve each equation. Verify the solution

$2x^2$ – 1 = 5 -> + 1 each side $2x^2$ = 6 -> divide 2 each side $x^2$ = 3 -> take the square root of each side x = $\sqrt{3}$ $2\sqrt{3}^2$ – 1 = 5 It is right. This example, a is 1. It is easy but If a is not 1, It will be very complex.

$\frac{-1x^2}{2}$ +6x -1 =0 -> Multiply each side by -2 $x^2$ – 12 + 2 = 0 -> half of the middle coefficient, then squared it $x^2$ – 12x + 36 – 36 + 2 = 0 -> factor the perfect square $(x-6)^2$ = 34 -> take the square root of each side x-6= +- $\sqrt{34}$ x= 6 ± $\sqrt{34}$

Using the Quadratic Formula to solve Quadratic Equation. In this chapter we have to know Formula. The formula is x= $\frac{-b+- \sqrt{b^2 - 4ac}}{2a}$ and a should be not zero. It can be used to determine the solution of any quadratic equation written in the form $latex ax^2 + bx + c = 0

$x^2$ + 2x + 1 = 0 x = $\frac{-2+- \sqrt{4+4}}{2}$ x = $\frac{-2+- 2\sqrt{2}}{2}$ x=-1+- $\sqrt{2}$

Week 5 blog post

Posted on October 7, 2018 by saejino2016

On this chapter, we learned Factoring Polynomial Expressions.

We learned important five words.

Common 15x+5xy -> 5x(5+y)

Difference of squares $x^2$ – 81 -> (x+9)(x-9)

Pattern $x^2$ +x+# $x^2$ +8x+12

product(12) sum(8)

1* 12 2*6
2* 6
3* 4

Easy $1x^2$

$x^2$ +17x+72 product(72) : 1*72, 2*36, 3*24, 4*18, 6*12, 8*9 sum(17) : 8*9 (x+8)(x+9)

Ugly $ax^2$

$5x^2$ -7x+2 product(10) : 1*10, 2*5 sum(-7) : 2*5 (5x-2)(x-1)

In Factoring Polynomial Expressions, there are some difficult case.

ex) $x^2$ +1.5x+0.5 -> $x^2$ + $\frac{15x}{10}$ + $\frac{5}{10}$

-> $\frac{10x^2}{10}$ + $\frac{15x}{10}$ + $\frac{5}{10}$

-> Find Common Factor $\frac{1}{2}$ ( $2x^2$ +3x+1) product(2) : 1*2 sum(3) : 1*2

-> $\frac{1}{2}$ (2x+1)(x+1)

In this case, we have to solve it this way. In this chapter, to find common factor is very important.

Week 4 blog post

Posted on September 30, 2018 by saejino2016

Sum this expression

$3\sqrt{5}$ + $4\sqrt{7}$ = $3\sqrt{5}$ + $4\sqrt{7}$

It can not be added, because it is same with x, y. Like 3x+4y. So it can not be added.

Sum this expression

$2\sqrt{3}$ + $6\sqrt{11}$ – $\sqrt{3}$ + $13\sqrt{11}$ = $\sqrt{3}$ + $19\sqrt{11}$

simplify this expression

( $3\sqrt{7}$ )( $2\sqrt{3}$ + $7\sqrt{13}$ ) = $6\sqrt{21}$ + $21\sqrt{91}$

On this chapter, I learned, the number in the root is same, it can be added and subtracted. But multiplication is possible to multiple that the number in root is not same. Also I learned new words. Like index, radicand, and radical.

$10\sqrt{15}$ In this root, index is 10, radicand is 15, and they are called radical.