anonymous one year ago Part A:The area of a square is (4x2 − 12x + 9) square units. Determine the length of each side of the square by factoring the area expression completely. Show your work. (5 points) Part B: The area of a rectangle is (16x2 − 9y2) square units. Determine the dimensions of the rectangle by factoring the area expression completely. Show your work. (5 points)

1. anonymous

@Mackenzie_Willa

2. anonymous

@welshfella

3. anonymous

I have the answer just not the steps! Part A:(2x -3)^2 Part B:(4x+3y)(4x−3y)

4. anonymous

@Nnesha

5. amilapsn

Do you know these? $a^2-b^2=(a-b)(a+b) \\ (a+b)^2=a^2+2ab+b^2$

6. anonymous

Because the first shape is a square, the length and the width must be the same. So the area must be a perfect square. We need to find an expression that , when multiplied by itself, will give $$4x^2-12x+9$$Start by taking the square roots of the first and last coefficients. Can you do that?

7. anonymous

Can you help me work it ut step by step plz

8. anonymous

*out

9. anonymous

What is the square root of the first term, $$4x^2$$?

10. anonymous

2x^2?

11. anonymous

Not quite. The square of 4 is 2, so you got that part right, but what is the square root of x^2? What can you multiply by itself to give x^2?

12. anonymous

*square root of 4

13. anonymous

I am confused

14. anonymous

Do you understand that$x \times x = x^2$

15. anonymous

yes

16. anonymous

OK. So the square root of $$x^2$$ is $$x$$. OK with that?

17. anonymous

oh okay.

18. anonymous

All right. So the square root of $$4x^2$$ is $$2x$$. In other words$2x \times 2x = 4x^2$Make sense now?

19. anonymous

Yes

20. anonymous

OK. So now you need the square root of the last term, +9. What is it?

21. anonymous

3

22. anonymous

Great. With these two square roots, you have determined that the two factors must be$4x^2-12x+9 = \left( 2x \text{ ? }3 \right)\left( 2x \text{ ? }3 \right)$All that is left is to choose the correct sign, plus or minus. For that, look at the second term in the given trinomial. What sign does it have?

23. anonymous

Minus.

24. anonymous

Perfect. Then that is the sign you use in the factors. Therefore,$4x^2-12x+9 = \left( 2x-3 \right)\left( 2x-3 \right)$and you have your answer.

25. anonymous

Keep in mind that this technique only works for factoring a perfect square trinomial.

26. anonymous

Are these the answers. I have been working them out too: Part A) 4x2 − 12x + 9 4x^2 -6x -6x + 9 2x( 2x - 3) -3(2x - 3) (2x - 3)(2x-3) length of each side is 2x-3 Part B) 16x^2 − 9y^2 (4x)^2 - (3y)^2 (4x - 3y) (4x+3y) dimensions are 4x-3y, 4x+3y.

27. anonymous

Yes, you are correct. Very well done!

28. anonymous

Thank you!

29. anonymous

You're welcome.

30. anonymous

@ospreytriple can you help with 1 more?

31. anonymous

An expression is shown below: f(x) = –16x2 + 60x + 16 Part A: What are the x-intercepts of the graph of the f(x)? Show your work. (2 points) Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points) Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

32. anonymous

Part A: To find the x-intercepts, you must solve the equation $$-16x^2+60x+16=0$$. Which method would you use; factoring by completing the square, or the quadratic formula?

33. anonymous

34. anonymous

OK. Go ahead. What do you get?

35. anonymous

Can you refresh me on the formula.

36. anonymous

For the general quadratic, $$ax^2 + bx + c=0$$, the x-intercepts are found using$x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$

37. anonymous

can you help me through this one

38. anonymous

I am really confused.

39. anonymous

Have you used the quadratic formula before?

40. anonymous

Just learning it.

41. anonymous

42. anonymous

Okay thank you! I do have 1 request can you say like part a and like part b and part c because i will get confused and don't know what to say.

43. anonymous

This is for Part A. If you look at the given trinomial and compare it with the general quadratic, you'll see that $$a=-16$$, $$b=60$$ and $$c=16$$. Putting these values into the quadratic formula, you have$x=\frac{ -b \pm \sqrt{b^2-4ac} }{2a } = \frac{ -60 \pm \sqrt{60^2-4\left( -16 \right)\left( 16 \right)} }{ 2\left( -16 \right) }$Can you complete these calculations?

44. anonymous

is it -5?

45. anonymous

One second. I'm working it out...

46. anonymous

okay ;]

47. anonymous

Wait is it... x = 4 and x = -1/4.

48. anonymous

Yayyyy! You got it. Those are the x-intercepts, $$x=4$$ and $$x=-\frac{1}{4}$$

49. anonymous

Yay :V. can you help with part B and C too?

50. anonymous

Part B: Vertex a maximum or minimum? What it means is, is the parabola opening upward or downward? If the parabola opens upward, then the vertex is a minimum. If the parabola opens downward, then the vertex is a maximum.|dw:1440091745676:dw| Do you know how to tell which direction the parabola opens?

51. anonymous

These are my answers so far are they correct? Part A.) f(x) = –16x^2 + 60x + 16 -4(x-4)(x+1/4) x-4=0 x+1/4=0 x=4 and x=-1/4 Part B.) vertex (15/8, 289/4) it is a maximum because the coeffficient of x^2 is -16 which is negative Part C.) You can graph the x intercepts and the vertex. That will give you the overall shape of the function.

52. anonymous

OK. That saves me a lot of work. Hope I'm not wasting your time. I see in Part A you factored instead of using the quadratic formula. Parts B & C are correct. Good job>

53. anonymous

Part c is correct?

54. anonymous

Yes.

55. anonymous

Oh. I wrote that down for notes rofl

56. anonymous

Thank you!

57. anonymous

You're welcome