Quadratic Functions

Lizbeth Acosta Week 4 Discussion

Solving Quadratic Equations

In this weeks discussion we are asked to solve two Quadratic Equations by using two main methods: factoring and Quadratic Formula. Two problems from our e-book Elementary and intermediate algebra (4th ed.) from pages 635 and 636. Both of our problems are based on the third letter of our last name. My last name is Acosta and the third letter is "O", therefore my problems are number 42 on page 635 and 52 on page 636.

page 635 # 42 I will be solving by factoring

y2-3y-10=0 given equation

(y2-5y+2y-10)=0 first use the technique of "completing the square" to rearrange the quadratic into the neat "(squared part) equals (a number)" format rewrite the middle term as a sum of the two terms

((y2-5y)+(2y-10))=0 Now group the first two terms together, then the second terms

(y(y-5)+2(y+5))=0 Find and factor the GCF for each group

(y+2)(y-5)=0 Factor the polynomial by grouping the first two terms together and find the GCF, next group the second two terms and find the GCF.

y+2=0

y-5=0 Solve for y

y - 2=0-2 Subtract 2 from both sides

y=-2

y-5=0 Add 5 to both sides

y+5=0+5

y=5 Combine both solutions

y=-2,5 Solution

page 636 # 52

x2-7x+4=0 given equation

x=-b+√b2-4(a)(c) where ax2+bx+c=0

2(a)

The Discriminant of a quadratic is the expression inside the radical of the quadratic formula

b2 – 4ac

a= 1, b= -7, c= -4 plug into equation

x=-(-7) ±√(-7)2-4(1)(-4) Simplify the section inside the radical...