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**ineedhelp****Member**- Registered: 2006-01-21
- Posts: 7

Do five out six, including Q2 and / or Q5. The problems that involve finding a solution (Q2, Q4, Q5) are to be done algebraically - by setting up and solving an equation. The object of Q2, for example, is not to 'play with' various consecutive-integer combinations until coming up with the solution (which is five and six) but to decide what the 'unknown' (x, say) should represent, and: (1) develop an equation that involves the unknown and utilizes the information provided; and then (2) solve that equation algebraically demonstrating that five and six are a solution by 'plugging in' these values into the equation is not solving it. Please show all work / explain your answers.

**1. Factor completely. 64x squared - 9**

I got 8x-9

**2. Solve the problem Find two consecutive integers such that the sum of their squares is 61. **

No answer. I have no idea.

**3. Factor 15z squared - 2z - 8 **

(-4, 2)

15 (z-4) (z+2)

**4. Find an integer solution to the following equation: (5x - 3) squared = 18x squared + 1 **

5x squared -9 = 18x squared +1

-13x squared -10

**5. Solve the problem. The printed matter on a 12 by 18 centimeter page of a book must cover 40 square centimeters. If all margins are to be the same width, how wide should the margins be?**

No answer. I have no idea.

**6. Factor the trinomial. x squared + 2xy - 24y squared **

(6, -4)

(x+6y) (x-4y)

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

1) It's (8x - 3)(8x + 3).

2) Let n be the number you're looking for. Then:

n² + (n+1)² = 61

n² + n² + 2n + 1 = 61

2n² + 2n - 60 = 0

2(n-5)(n+6) = 0

n = -6 and n = 5.

So, both -6, -5 and 5, 6 will work.

3) (3z+2)(5z-4)

4) (5x-3)² = 18x² + 1

25x² - 30x + 9 = 18x² + 1

7x - 30x + 8 = 0

(x-4)(7x - 2) = 0

x=4 and x = (2/7)

The integer solution is x=4.

5) The total area of the page is 12*18 = 216 cm². The printed material covers 40 cm², leaving 176 for the margins. Set up an equation for the margins and solve it.

Let w = the width of the margins. Then,

12w + 12w + 2(18-2w)w = 176

-4w² + 60w - 176 = 0

-4(w-11)(w-4) = 0

w = 11 or w = 4

Two widths would work, but 11 is almost 12, which is one of the dimensions of the page, so that's probably not the best choice. Let's check it with 4:

48 + 48 + 40 + 40 = 176

6) You're correct on this one.

Note that I used a calculator to factor some of those quadratics, since factoring isn't one of my strong points. Lacking that, I'd resort to the quadratic formula.

Note also that when factoring, I always have to expand my factored expression to check and see if it's right.

*Last edited by ryos (2006-02-11 19:19:07)*

El que pega primero pega dos veces.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

No calculator needed thanks to a lesson given by Mathsyperson earlier in this forum. In any factorizable quadratic we can break up the b constant into two terms.

ax² + bx + c = 0 also equals ax² + dx + ex + c = 0

where d and e will be the factors of ac that also satisfy d+e = b

For the problems 1, 3, and 6;

1) 64x² - 9

ac = -576 and d + e = 0, the only possible factors are ±√576 or 24 and -24

So our original equation becomes;

64x² - 24x + 24x - 9

Factoring the first two terms and the second two terms gives;

8x(8x - 3) + 3(8x - 3)

Which simplifies to;

(8x + 3)(8x - 3)

3) 15z² - 2z - 8

ac = -120 and d + e = -2, the only two factors that solve this are -12 and 10, just plug them in and factor as above.

15z² - 12z + 10z - 8

3z(5z - 4) + 2(5z - 4)

(5z - 4)(3z + 2)

6) This is treated the same as those above, the extra variable does not change the method.

x² + 2xy - 24y²

ac = -24y and d + e = 2y, only 6y and -4y can do that so;

x² + 6xy - 4xy - 24y²

x(x + 6y) - 4y(x + 6y)

(x + 6y)(x - 4y)

Mathsyperson explained this technique flawlessly in an earlier thread within this forum, I believe that it was "factoring quadradics" by rickyswaldio. My memory is slightly faded, but that is where I think it was posted before.

*Last edited by irspow (2006-02-12 07:30:00)*

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