Methods of Solving Quadratic Equations In this activity, you will choose a problem-solving method and use it to find the solutions to quadratic equations. Question 1 Part A Question Type the correct answer in each box. Write your answers in decimal form, rounded to the nearest tenth, if necessary. Type the solution with the smaller value in the first blank. (Hint: to complete your calculations, you may need to use mental math.) Select and use the most direct method to solve 2x(x + 1.5) = -1. x = or x =

step-by-step explanation:

quartic functions has x = –1 and x = –2 as its only two real zeroes:

f(x) = a(x+2)(x+1)   a: real number

4/6 = 2/3

so it should take 3/3 hours per man. so 9 hours is the answer 3*3 =9

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3