algebra experts needed!

[buddy]

Here's a few ninth grade quadratic polynomials which need factored. I'm trying to teach my daughter to handicap and she doesn't have time for this algebra stuff.

For each of the following quadratic functions, use the quadratic formula or factoring to find the x-intercepts and the factors. For each of the quadratic equations, find the roots(solutions).

Quadratic Formula = Ax^+Bx+C

1. y = 6x^-29x+5 = ( ) ( )

2. 3x^-4x-24=0

3. -9x^+27x-36= ( ) ( )

4. y+10x^+14x-12= ( ) ( )

5. 3x^-3x+2=0

6. y=6x^-9x+2 = ( ) ( )

Thanks,
Buddy & Kid

 

[TORONTO-VIGILANTE]

funny line du jour: part deux

I'm trying to teach my daughter to handicap and she doesn't have time for this algebra stuff.

:lol:

 

[Evil Genius]

This is what you need to find the answers you seek. This is how you solve an unfactorable polynomial.

<img src=http://www.sosmath.com/algebra/quadraticeq/quadraformula/img9.gif>


Since I'm a nice guy here's the answers:
<font size=-2 color =red> (any number)^.5 MEANS SQAURE ROOT</font>

1. (29 +or- (721)^.5 ) / 12

x = 4.65428693 or .179046403

 

[Evil Genius]

number 2

(4 +or- (304) ^ .5) ) / 6

which can be reduced to

(2 +or- (2*(19) ^ .5) ) /3



so x = 3.572599296 or -2.239265962

 

[Evil Genius]

number 3


-9x^2 + 27x -36

can be rewritten as

x^2 -3x +4


you get (3 +or- (-7)^.5 ) / 2


since you cannot solve the square root of a negative number, at least not in 9th grade, there is no solution.

This means that there is no x-intercepts---it is a parabola opening up going through the point y=4 x=0

 

[Evil Genius]

number 4


you have to get it in the form of y = ax^2 + bx + c first so you get......



y = -10x^2 -14x +12 which after dividing by -2 is equal to


y = 5x^2 + 7x -6

-- this can be factored to (5x - 3) (1x + 2)

x= 3/5 & -2

 

[Evil Genius]

number 5


Five is the same as 3. You get.....

(3 +or- ( (-15)^.5) ) / 6

square root of -15 does not exist so there are no x intercepts (which means when y=0 x does not exist)


This is another parabola opening up going through the point y=.66666 x=0

 

[KotysDad]

number 3

since you cannot solve the square root of a negative number, at least not in 9th grade, there is no solution.

This means that there is no x-intercepts----which is a straight horizontal line going through the point y=4 x=O

Evil,

Can you explain this last statement?? This function is just a parabola that opens up and sits above the x-axis.

What is the straight horizontal line for?? It isnt an asymptotic function.

 

[buddy]

3. -9x^+27x-36=y

^ = squared

so....

3. -9x squared + 27 x - 36 = y

answer: 9 (-x + 4) (x -1) = y

is this right?

 

[KotysDad]

3. -9x^+27x-36=y

^ = squared

so....

3. -9x squared + 27 x - 36 = y

answer: 9 (-x + 4) (x -1) = y

is this right?

Buddy,

No, that one doesnt factor into integer solutions.

 

[Evil Genius]

Evil,

Can you explain this last statement?? This function is just a parabola that opens up and sits above the x-axis.

What is the straight horizontal line for?? It isnt an asymptotic function.

Good point. Sorry it has been a long while. KotysDad is right. There is no x intercept. It is just a parabola which doesn't cross the x axis.

This also applies to number 5.

 

[Evil Genius]

number 6


you get (9 +or- (33)^.5) / 12


so x = 1.228713554 or .2712864461

 

[Evil Genius]

Hey buddy,

I fixed numbers 3 and 5 because they were wrong. Thanks to Kotysdad for pointing out my errors.


Before I said they were straight lines, but they aren't, they are parabolas (bell shaped curves) that do not cross the x axis.



I hope this helps your daughter with her homework....maybe someday she can give me a freakin winner in football. LOL