Can Some One Help me My Assignment please check these attachment and answer the questions # 22 23 24 25 26 27 28 

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jerichorayel | College Teacher | (Level 2) Senior Educator

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22. x^2 + 2x -3 =0

(x -1 )(x +3) = 0

the roots are:

x = 1

x = -3

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23. 3b^2 - 14b + 8 = 0

(3b -2)(b - 4) = 0

b = 2/3

b = 4

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24. x = -2 and x = -7

x + 2 =0 ; x + 7 =0

(x+2)(x+7) = 0

x^2 + 2x + 7x + 14 = 0

x^2 + 9x + 14 = 0 -> quadratic equation

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jerichorayel | College Teacher | (Level 2) Senior Educator

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25.-5x^2 + 2 = 3x + 2  -> arrange the equation in the standard quadratic form

5x^2 + 3x + 2 - 2 = 0 

5x^2 + 3x  = 0 

Using the quadratic formula, the roots are:

x = -3/5 

x = 0

You can verify the answer by substituting the values of x.

5x^2 + 3x  = 0 

5(-3/5)^2 + 3(-3/5)  = 0 

9/5 + (-9/5) = 0

0 = 0 check

and

5(0)^2 + 3(0)  = 0

0 = 0 check

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26. let the two numbers be x and y.

x + y = 108                 equation 1

x*y = 2592                  equation 2

Transform equation 1 into:

y = 108 - x                  equation 3

Substitute equation 3 into equation 2:

x(108 -x) = 2592

108x - x^2 = 2592 

arrange the equation into standard quadratic equation:

x^2 - 108 + 2592 = 0

(x - 36) (x - 72) = 0

x = 36

x = 72

Substituting the values of x in the equation 1 would give

x + y = 108 

36 + y = 108

y =108 -36 = 72

OR

x + y = 108 

72 + y = 108

y = 36

The two numbers are 36 and 72. 

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28. The correct way in solving for the root is:

9y^2 - 81y = 0

9y (y -9) = 0

y = 0

y = 9

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steamgirl | Student, College Junior | (Level 1) Honors

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22) a root is wherever the function crosses the x-axis or where y=0. In this case -3 & 1.

23) Most people find the diamond method useful for theses problems. Basically you find which two numbers multiply to give you the c term and add to give you the b term in a standard equation.

ax +bx+cx=0 

In this case you have an a term and you have to take that into consideration. I would use the reverse box method. Set up a box, like when you would combine instead of simplify and work backwards.

In this problem, you end up with (b-4)(3b-2).

24) Your looking for where y=0 and x is both -2 & -7. Therefore, set x= to these numbers individually and solve for 0. This will give you x+2 & x+7, just multiply out to find your standard equation.

25) First put the equation in standard form: ax+bx+cx=0. Then plug into the equation -b +\-( √b^2-4ac)\2a, with the letters matching from the equation. You will end up with one being zero, the other can be plugged into the original equation and will equal zero.

26) Give these unknown # the value x and y. x+y = 180 & xy=2592, use one of the equations to solve for a variable and plug into the other. Such as x=180-y, then (280-y)y= 2592. And solve.

27) This problem says to use technology to solve, which usually means a calculator, the only way around this is to plot the graph yourself, you'll want to make a table. This will actually be a nice graph since common sense says that height and time must start at zero. The end of the graph is when the ball hits the ground again, and since there is no negative distance it is when y is back at zero. 

The domain of a graph is all the x values the function covers. We know that it's lowest x is zero, the highest is wherever y is zero at the end of the graph. The range is the same concept except for the y values. Wherever x reaches it's max is the highest y value. When you find these write them with greater and less than signs for your final answer.

28) The only thing wrong with this solution is that they had an extranx still in the ().

Should have been: 9y(y-9). From there just solve.

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rachellopez | Student, Grade 12 | (Level 1) Valedictorian

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22. A root of a polynomial is what x would equal to make the whole equation equal zero. To get the roots of the polynomial x^2+2x-3 you can simply factor to get (x+3)(x-1) and then set both factors equal to zero.

x+3=0                    x-1=0

x=-3                       x=1

You can also look at the graph you were given and the roots would be the points the parabola touches the x-axis.

23. 3b^2-14b+8=0

First, in order to get 3b^2 you need to multiply 3b and b. (3b   )(b    )

Now you have to think what 2 numbers will multiply to be 8 but add to get -14. Keep in mind when you distribute you will be multiplying the second number by 3 as well. (3b-2)(b-4)

Once you have your factors you just set both equal to zero and solve.

b=2/3                   b=4

24. When you are given your solutions of x=-2 and x=-7 you just have to work backwards.

x=-2                    x=-7

x+2=0                 x+7=0

Your factors are (x+2) and (x+7). Now you have to distribute, giving you    x^2+9x+14

25. To do the quadratic formula you have to set your equation equal to zero and then label your equation with a, b, and c.

-5x^2+2=3x+2

-5x^2-3x=0

a=-5                b=-3         c=0

`(3+-sqrt((-3^(2))-4(-5)(0)))/(2(-5))`

x= -3/5          x=0

26. x+y=108                xy=2592

Make y be by itself in the first equation which gives you y=108-x. Now substitute y in the second equation for 108-x

x(108-x)=2592

-x^2+108x=2592

-x^2+108x-2592=0

You can either factor at this point or use the quadratic formula. Either way x=36                x=72

27. In order to find the maximum height you have to put your equation into vertex form. To do this you have to use the complete the square method.

y+___=-4.9x^2+10x+____

y+___=-4.9(x^2+(10/-4.9)+___)

I'm not really sure where to go from here, but I hope that helps you start.

28. When you solve for a root you have to set the whole equation equal to zero and either factor or use the quadratic formula. What they did wrong is when they factored out the 9y they would be left with 9y(y-9) not 9y(y^2-9). The correct way would be

9y^2-81y=0

9y(y-9)=0

9y=0               y-9=0

y=0                 y=9

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