PLEASE HELP ME! math FINAL tomarow help ASAP?

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

For the quadratic equation ax^2 + bx + c = 0, the key thing to check is the discriminant, Δ = b^2 - 4ac.
If Δ is positive there are two distinct real roots.
If Δ is zero there is a single repeated real root.
If Δ is negative there are no real roots.

25c^2-20c+4=0: discriminant is (-20)^2 - 4(25)(4) = 400 - 400 = 0, so there is a single repeated real root.

x^2-12x+k=5: discriminant is (-12)^2 - 4(1)(k) = 144 - 4k. For equal roots we need 144 - 4k = 0 => k = 36.

y^2+ky+36=0: discriminant is k^2 - 4(1)(36) = k^2 - 144
= 0 if k = ±12.

2y^2+by-12 = 0: one root is 4, so we know
2(4)^2 + 4b - 12 = 0
=> 4b = -20, b = -5
So the equation is 2y^2 - 5y - 12 = 0, and we know one factor is (y - 4), so we can factorise it as
(y-4) (2y+3) = 0
So the other root is -3/2.

If the roots are α and β, we can write the equation as
(x-α) (x-β) = 0
<=> x^2 - (α+β) x + αβ = 0

This gives us an alternate way of solving the previous question: 2y^2+by-12 = 0 <=> y^2 + (b/2) y - 6 = 0, so the product of the roots is -6. Since one root is 4 the other root must be -6 / 4 = -3/2.

If α+β = 3 and αβ = 4.5, the equation is
x^2 - 3x + 4.5 = 0
or equivalently 2x^2 - 6x + 9 = 0.

α+β = -4 and αβ = 5
=> equation is x^2 + 4x + 5 = 0

2x^2 - 5x - 3 = 0
<=> x^2 - (5/2) x - (3/2) = 0
=> sum of roots is 5/2 and product of roots is -3/2
So if we add these together we get 1.


Summary of Properties:
For the quadratic equation ax^2 + bx + c = 0
* The discriminant Δ = b^2 - 4ac determines the nature of the roots.
- - If Δ > 0 there are two distinct real roots.
- - If Δ = 0 there is a single repeated real root (or two equal real roots, same thing).
- - If Δ < 0 there are no real roots (there are two complex conjugate roots, but I don't think you're up to that yet.).

* The sum of the roots is -b/a and the product of the roots is c/a.

* If r is a root, then ar^2 + br + c = 0 (this is what it means to be a root) and (x - r) is a factor of ax^2 + bx + c.

Good luck!

Answer 2

I'm so sorry but if i was smarter i could help you a lot, but i know your properties: communitive, associative, and distributive. What grade are you in???

Answer 3

You are soooooooo screwed

Answer 4

i forgot how to do this.....srrry. good luck!