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
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
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.
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.
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)