PLEASE HELP!!! math b final TOMAROW please help nowww?

Question

ok heres some quentions from the review packet I have no idea about... please help answer them and explain....
*determine the nature of the roots of the quadratic equation 25c^2-20c+4=0
*find the value of k that will make the roots of the equation equal: x^2-12x+k=5
*find the values of k wich will make the roots of the equation equal: y^2+ky+36=0
*if one root of the equation 2y^2+bx-12=0 is 4, find the other root.
*for wich equation does the sum of the roots equal 3 and the product equal 4.5?
*write an equation in the form ax^2+bx+c=0 if the sum of the roots is -4 and the product is 5
*if the sum of the rootsof the equation 2x^2-5x-3=0 is added to the product of the roots,the result is...
* also please include any propertys you know, they will be on the test but i dont have the review sheet on that because i was sick
thank you sooooooooooo sooooo much!!!!!

Answer 1

Given an equation dx^2 + ex + f = 0,
x = (-e +/- sqrt(e^2 - 4 * d * f))/(2 * d)

25c^2 - 20c + 4 = 0
d = 25
e = -20
f = 4

c = (20 +/- sqrt(400 - 4 * 25 * 4))/(2 * 25)
c = (20 +/- sqrt(400 - 400))/50
c = 20/50 = .4



To make the roots of the equtation equal, it must be a perfect square. For example, in (x^2 + 6x + 9), x = -3.

x^2 - 12x + k - 5 = 0
. . 12/2 = 6
. . 6^2 = 36
k - 5 = 36
k = 41



k/2 = sqrt(36).
k/2 = +/- 6
k = +/- 12



I'm assuming this is 2y^2 + by -12 = 0

If a root is 4, then it factors to (y - 4).

The other factor must be something (2y + 3), so that 3 * -4 = -12, and 2y * y = 2y^2. Then by = -8y + 3y = -5y
b = -5



x + y = 3
xy = 4.5
x = 4.5/y
4.5/y + y = 3
4.5 + y^2 = 3y
y^2 - 3y + 4.5 = 0
Note that these are as above, but reverse, because when the root is x = a, the factor is (x - a).


x^2 - 5x + 4 = 0, as above.



x = (5 +/- sqrt(25 + 24))/4
x = (5 +/- 7)/4
x = 3 or -.5

Sum = 2.5
Product = -1.5
Total = 1

Answer 2

*determine the nature of the roots of the quadratic equation 25c^2-20c+4=0

You'll first have to put this into a(x-h)^2+k form. Before we do this, we need to take out that 25 in front of c^2:

25(c^2 - 4/5c + 4/25) = 0

Now complete the square, when finished, you'll get:

25[(c - 2/5)^2 + 16/25] = 0

So, the vertex is (2/5 , 16). The root is shifted 2/5 to the right, and up 16. The minimum value is y >= 16.

Sorry, that's it for tonight. I just came back from tutoring math and my head hurts. Hope at least you're cleared on one question! Good luck on your test tmw!

Answer 3

When they mean root, they are talking about the zeros that cross
the x-axis of your graph. Anyway, For the first one, your roots are
2/5 and 2/5 because it was a perfect square binomial. Just solve for K to get your possible values(s). For the second one, Just factor to find the roots, set each one equal to zero and solve. Ex:

25c^2 - 20c + 4 = 0. (factor)

(5c-2)^2 = 0.

5c-2 = 0 5c-2 = 0

x = 2/5, 2/5. These are your roots. For the sum of the roots, you might get different answers.

Answer 4

1) *determine the nature of the roots of the quadratic equation 25c^2-20c+4=0....

25c^2-20c+4=0
25c^2-10c-10c+4=0
5c(5c-2)-2(5c-2)=0
(5c-2)(5c-2)=0


for wich equation does the sum of the roots equal 3 and the product equal 4.5...

x^2+3x+4.5
(x+1.5) (x+1.5)

Answer 5

use this formulation destructive b plus or minus the sq. root of b^2 - 4ac throughout 2a i desire u get it! it is observed as the quadratic formulation sum of roots: -b/a made from roots: c/a sum over made from roots: -b/c