# Need help with lots of different types of problems

1. Sep 7, 2006

### RubyRed

I already posted a thread about a "physics" (although I am actually in basic college algebra) but I have a few more that I need help with.

1) A jumbo chocolate bar with a rectangular shape measures 12 cm in length, 7cm in width and 3 cm in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 20%. To accomplish this reduction, management decides that the new bar should have the same 3 cm thickness but the length and width of each should be reduced an equal number of centimeters. What should the dimensions of the new candy bar be?

~Now I know that the formula for volume is lwh. So I took the original measurements 12 x 7 x 3 and got 252. I reduced that by 20% which would be 252-50.4= 201.6 to get the new volume of the reduced bar. Now I know that the thickness is going to stay the same so I think the new equation would be 201.6=lw x 3. I need to make sure that the length and width get reduced the same amount so I have to make sure that they stay 5 cm apart (if original l=12 and w=7 12-7=5cm) so I made the equation say 201.6= w+5 x w x 3 (x's are multiplication symbols by the way) but now what do I do..do I subtract 5 from both sides and make the w's w^2 and then divide both sides by 3 and square root the answer to get w? And then once I know that I can find l? I don't know what to do from here!

2)Find k such that the equation x^2 - kx +4 =0

Ok so I subtacted 4 from each side to get x^2 - kx= -4 Now I am horrible at these so I am probably way off but then I divded both sides by x....so that I had x-k = -4 over x (I don't know how to type fractions, sorry!) but I had no idea what to do from there...I am supposed to only keep a constant on the right correct? So moving x over seemed wrong to me, but with only one given number in an equation I don't know what to do.

I guess that is it for now, I have more but maybe I can do them once I know how to do these ones. Thanks guys!

2. Sep 7, 2006

### bob1182006

#1 has me a bit stumped at the moment :s
but #2 you did good so far getting $$x-k=\frac{-4}{x}$$ and now you just to have to continue solving for k by having k by itself on one side of the equation

3. Sep 7, 2006

### RubyRed

I just realized that I didn't type the entire problem for number 2....it should say "find k such that the equation x^2-kx+4=0 has a repeated real solution." So I should move x over in which case Iwould have -k=-4-x....which would make k=4+x and that is as far as I can go right? But what about the repeated real solution part that I am supposed to find??? Kinda confusing to me

4. Sep 7, 2006

### bob1182006

uh repeated real solution? never heard of that sorry. Does your textbook have any mention of repeated real solution?

but in the solution $$k=4+x$$ you droped the /x on the 4

5. Sep 7, 2006

### RubyRed

In my text book all it says is "When the left side factors into two linear equations with the same solution, the quadratic equation is said to have a repeated solution." It doesn't say the word real in the definition but I am assuming they are referring to a repeated solution with real numbers....still a bit confusing to me. ohh so the solution should be k= 4/x + x , can you add that together and make it k=4/2x which would just = 2x right? so is k=2x the answer?

6. Sep 7, 2006

### bob1182006

ok finally i solved #1 :s so i can help you out with it:

so you know from the question that 12 and 7 will be lowered by the same amount right? so one way to solve it is to create 3 equations to solve this problem:

the original

l*w*3=201.6
$$l=12-x$$
$$w=7-x$$

so ill let you have a go at it and see if you can solve it

7. Sep 7, 2006

### RubyRed

ok can you give me the next step to get started on because I am stumped...

8. Sep 7, 2006

### RubyRed

wait! ok so 201.6 = (12-x)(7-x)(3) is that the right direction?

9. Sep 7, 2006

### bob1182006

ooo i get that now, what they mean is like in the quadratic equation a (real) repeated solution would be when the determinant $$b^2-4ac=0$$ which would give a repeated solution, what they also want is this to be a real solution, no negative square roots.

so using the original equation you just plug in for b, a, c and solve for k

just in case you forgot the quadratic equation is in form $$ax^2+bx+c=0$$

10. Sep 7, 2006

### RubyRed

so should I foil the first 2 and then multiply by 3? ugh but then how do I separate what is l and what is w?

11. Sep 7, 2006

### bob1182006

yes that's the right direction, but i recomend before plugging in the value's for l and w to move the 3 to the right side to make things a bit easier, and then do foil and solve for x

Last edited: Sep 7, 2006
12. Sep 7, 2006

### RubyRed

so after foiling 201.6= 252-57x+3x^2

that doesn't seem right to me... ok I am confused again

13. Sep 7, 2006

### bob1182006

try dividing both sides by 3, and then foiling, then you'll have a quadratic equation which you can solve using the quadratic formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

or you could do that now by moving the 201.6 over to the right side which would give you
$$3x^2-57x-50.4=0$$

Last edited: Sep 7, 2006
14. Sep 7, 2006

### RubyRed

so if I divide both sides by 3 I have 67.2 = (12-x)(7-x)/3 is that right so far?

15. Sep 7, 2006

### RubyRed

but how do I foil it if it is all over 3?? or when I divide by 3 does it just cancel out the (3) and not affect the others. Should it be 67.2=(12-x)(7-x)

16. Sep 7, 2006

### bob1182006

no, if you divided by 3 you'd just have 67.2=(12-x)(7-x)

17. Sep 7, 2006

### RubyRed

ok I thought you had to divide every single thing by 3...my bad(jeepers it has been so long since I have done this!), ok so now I have 67.2=84-19x+x^2 so now I can plug it into the equation you showed me?

18. Sep 7, 2006

### RubyRed

19. Sep 7, 2006

### bob1182006

well if you had it lw3 then it would be 3l*3w, so dividing by 3 would just leave lw or (12-x)(7-x)

edit: sorry didnt see your last posts

kz you got everything right so far, now just move the 67.2 over to the right side so its all =0 and plug a/b/c intot he quadratic formula to solve for x

Last edited: Sep 7, 2006
20. Sep 7, 2006

### RubyRed

ok I got x=12 or x=7 but those were the original 2 numbers. lol...how did i get that? I will try to show what I did. I foiled and got 67.2= 84-19x+x^2 I plugged it in and got x= -(-19) +- square root of 19^2-4(1)(84) over 2(1) (you might want to write that out to visualize it) so then I got it down to x=19+- square root of 361-336 over 2. Which = x=19 =- square root of 25 over 2, which is then x=19 +- 5 over 2. So then I did x=19-5 over 2 which is x=14 over 2 which is x=7. Then 19+5 over 2 which is x=24 over 2 which is 12....but those can't be right becuse then it would be the SAME EXACT as before I reduced it by 20%....how did that happen?