Week 4 discussion MAT 222
Since the third letter of my last name is the letter L, I have to solve problem 46 on page 636 and problem 48 on page 646.
This first problem will be solved by factoring the polynomial formula of ax^2+bx+c to get a*c(-30) and to add up to (-1) we have a=2,b=-1,and c=-15=0. The new equation is (2m^2-m-15)
Original equation: 2m^2-m-15=0
Step 1. (2m^2-6m+5=15=0) To start we separate the fraction. Then solve both sides of the original equation separately
Step 2. 2m(m-3)+5(m-3)=0 Move both factors to the left of the 0 a) M-3=0
Step 3. 2m+5=0 Divide each term in A by “2” Solve: a) m=-5/2
For the second problem the quadratic formula will be used to solve the equation. Since a=2y^2 b=3y c=6 and we know these are integers and two terms complete the square, we can also see in our equation we have the discriminant so we determine if we have a prime polynomial.
Using the quadratic formula: Original equation: 2y^2-3y-6=0
Step 1: y=-b± b^2-4ac/(2a) so we will have our equation of ay^2+by+c=c
Step 2: y= (-3) ± (-3) ^2-4 (2) (-6) /2 (2) We plug in our terms
Step 3: y= 3± (-3) ^2-4 (2) (-6) /2 (2) We *by -1 to our terms outside the radical then do the math and simplify inside the radical
Step 4: y=3 +57 /2 (2) Simplify the denominator.
Step 5: y=1/4 (3+ 57 ) Next we need to solve for ±
Step 6: 1/4 (3 - 57 ) Next we solve for –
Solve: 1/4 (3 + 57,¼ (3-57) The final answer gives us the following result: y≈2. 6375, -1.1375 2.638,-1.138
This first problem will be solved by factoring the polynomial formula of ax^2+bx+c to get a*c(-30) and to add up to (-1) we have a=2,b=-1,and c=-15=0. The new equation is (2m^2-m-15)
Original equation: 2m^2-m-15=0
Step 1. (2m^2-6m+5=15=0) To start we separate the fraction. Then solve both sides of the original equation separately
Step 2. 2m(m-3)+5(m-3)=0 Move both factors to the left of the 0 a) M-3=0
Step 3. 2m+5=0 Divide each term in A by “2” Solve: a) m=-5/2
For the second problem the quadratic formula will be used to solve the equation. Since a=2y^2 b=3y c=6 and we know these are integers and two terms complete the square, we can also see in our equation we have the discriminant so we determine if we have a prime polynomial.
Using the quadratic formula: Original equation: 2y^2-3y-6=0
Step 1: y=-b± b^2-4ac/(2a) so we will have our equation of ay^2+by+c=c
Step 2: y= (-3) ± (-3) ^2-4 (2) (-6) /2 (2) We plug in our terms
Step 3: y= 3± (-3) ^2-4 (2) (-6) /2 (2) We *by -1 to our terms outside the radical then do the math and simplify inside the radical
Step 4: y=3 +57 /2 (2) Simplify the denominator.
Step 5: y=1/4 (3+ 57 ) Next we need to solve for ±
Step 6: 1/4 (3 - 57 ) Next we solve for –
Solve: 1/4 (3 + 57,¼ (3-57) The final answer gives us the following result: y≈2. 6375, -1.1375 2.638,-1.138