math squad! girl in need of assistance?

Question

i'm terrible at math...no matter how much i try i always get it wrong...so yahoo answers is really my last resort. im so much better at science, but unfortunately my quiz isnt in science today :(....if you can help me out here then please please do.

1. Solve the equation and select the letter of the correct answer.

0 = k^2 – 12k + 35
The solution set is {5, 7}.

The solution set is {–5, –7}.

The solution set is {5, –7}.

The solution set is {–5, 7}.


3. Solve the equation and select the letter of the correct answer.

u^5 – 13u^3 + 36u = 0
The solution set is {0, 2, 3}.

The solution set is {–3, –2, 0}.

The solution set is {–3, –2, 2, 3}.

The solution set is {0, –2, 2, –3, 3}.



5. Find an equation in standard form with integral coefficients that has the given solution set.

{–1, 5}

x^2 – 4x – 5 = 0

x^2 + 4x – 5 = 0

x^2 – 4x + 5 = 0

x^2 – x + 5 = 0

Answer 1

PROBLEM 1:

You can approach this one of two ways. Either you can try plugging in the answers to see if you get zero, or you can try and factor.

If you try the first approach, you find:
0 = 5² - 12(5) + 35
0 = 25 - 60 + 35
0 = 60 - 60

0 = 7² - 12(7) + 35
0 = 49 - 84 + 35
0 = 84 - 84

So the answers are {5, 7}

Using the factoring method, you get:
x² - 12x + 35 = 0
(x - 5)(x - 7) = 0

That leaves two solutions:
x - 5 = 0
or
x - 7 = 0

Solving:
x = 5
or
x = 7

This is the same as the first answer {5, 7}

PROBLEM 2:

Same methods will work. I'd use factoring:

Factor out a common 'u' first:
u(u^4 - 13u² + 36)

Notice you have only even powers in the parentheses, so temporarily replace u² with k...
u(k² - 13k + 36)

This can be factored easily as:
u(k - 4)(k - 9)

Put u² back in for k:
u(u² - 4)(u² - 9)

The parentheses are each a difference of squares, so use the rule for that (a² - b²) = (a - b)(a + b)

u(u - 2)(u + 2)(u - 3)(u + 3) = 0

Now you have your *five* solutions:
u = 0
u = 2
u = -2
u = 3
u = -3

{0, -2, 2, -3, 3}

PROBLEM 3:

Start with your two solutions
x = -1
or
x = 5

Now put everything on one side (add 1, subtract 5)
x + 1 = 0
or
x - 5 = 0

Now you can multiply these together without affecting things:
(x + 1)(x - 5) = 0

Then expand out with FOIL:
x² - 4x - 5 = 0

Answer 2

k^2 – 12k + 35 = 0

(k - 7)(k - 5) = 0 {factor the trinomial}

Statement is true if when either factor = 0

k - 7 = 0
k = 7

---

u^5 – 13u^3 + 36u = 0

u(u^4 - 13u^2 + 36) = 0 {common factor}

u(u^2 - 4)(u^2 - 9) = 0

The binomials are each differences of two squares like a^2 - b^2 whose factors are (a + b)(a - b)

u(u + 2)(u - 2)(u + 3)(u - 3) = 0

Set each factor to 0

u = 0, -2, 2, -3, 3

---

5. Find an equation in standard form with integral coefficients that has the given solution set.

{–1, 5}

The factors would be

(x + 1)(x - 5)

x^2 - 5x + x - 5

x^4 - 4x - 5

k - 5 = 0
k = 5

Answer 3

These all have to do with factoring polynomials

0 = k^2 – 12k + 35
0 = (k - 5)(k -7) [numbers that multiply to 35 but add to -12]
k=5 or k=7
{5,7}

u^5 – 13u^3 + 36u = 0
u(u^4 - 13u^2 + 36) = 0
u(u^2 - 9)(u^2 - 4) [numbers that multiply to 36 and add to -13]
u(u+3)(u-3)(u+2)(u-2) = 0
u = {0,-3,3,-2,2}

(x+1)(x-5)=0
x^2 -4x -5 = 0

Answer 4

0 = k^2 – 12k + 35

0 = (k-7)(k-5)

so k-7 = 0 or k-5 = 0, this means k=7 or k=5 (5,7) is the answer.

u^5 – 13u^3 + 36u = 0

first factor out a "u"

u(u^4-13u^2+36)=0

u(u^2-9)(u^2-4)=0

so u=0 or u^2-9=0 or u^2-4=0, so u=0 or u=sqrt(9) or u=sqrt(4), so u=0,3,-3,2,-2

this first equation (x^2-4x-5) is the answer to #5

Answer 5

Answering any of those questions would be cheating...

Don't you have a math coach that you can go to? That way, you'll learn how to answer those questions in the future. Instead of getting them (answers) from yahoo.

Answer 6

hey your on K12 right? well if you email me I bet my brother can help you in your math cause I think your in his class. And your in my English so then maybe we can both help each other out ^.^
so email me if you please and ill be happy to add you to the circle XP "hope ya see this"