equation | Saejin's Blog

On this week, we learned how to graph Linear inequality. First, we have to check y-intercept and slope. and next, we have to check inequality sign. If any equation has equal sign, we have to draw a solid line, however the equation doesn’t have equal sign, we have to draw broken line. And if we graph the linear inequality, the graph of the equation divides the coordinate plane into two region, and to put the test number into the equation. and if the equation include the part of to include the test number, we have to paint that part.

Ex) Determine whether each point is a solution of the given inequality

4x-3y ≥ 6 A(4,3) 4(4)-3(3) ≥ 6 -> 16-9 ≥ 6 -> 16 ≥ 15 This point is solution.

7y+2x < -10 B(-8,1) 7(1)+2(-8) < -10 -> 7-16 < -10 -> -9 < -10 This point is not solution

-y+5x ≤ 17 C(2,-7) -(-7)+5(2) ≤ 17 -> 7+10 ≤ 17 -> 17 ≤ 17 This point is solution

-3x+8y+5 > 0 D(-5,-3) -3(-5)+8(-3)+5 > 0 -> 20-24 >0 -> -4 > 0 This point is not solution

Also on this week, we learned how to graph Quadratic inequality. First, we have to find the coordinates of vertex, slope(a), x-intercepts, y-intercept,and direction of opening. And check the inequality sign of the quadratic equation. If the equation has equal sign, we have to draw a solid curve, if the equation doesn’t have equal sign, we have to draw a broken curve. And next, to put the test number into the equation.(Test points can be used to determine which region satisfies the inequality.) And also we have to paint the part of to include the test number.

Ex) To graph Quadratic inequality

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+- $\sqrt{2}$

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