- #1

yungman

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I work to this point and get stuck:

https://www.physicsforums.com/attachments/270843

Can anyone help me?

Thanks

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- Thread starter yungman
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- #1

yungman

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I work to this point and get stuck:

https://www.physicsforums.com/attachments/270843

Can anyone help me?

Thanks

- #2

mfb

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- #3

yungman

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I really don't know, this is a question from my grand daughter and this caught me cold. I am ashamed!

I set the equation to zero to get two roots.

- #4

Mark44

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- #5

SammyS

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This appears to be from a set of exercises for a lesson on factoring.I have problem solving this. The question is #19 in the first attachment. My work is in the second attachment

View attachment 270842

I work to this point and get stuck:

https://www.physicsforums.com/attachments/270843

Can anyone help me?

Thanks

There are several methods for factoring a trinomial such as this. What method or methods have your granddaughter studied?

- #6

yungman

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I have to get hold of her. thanksThis appears to be from a set of exercises for a lesson on factoring.

There are several methods for factoring a trinomial such as this. What method or methods have your granddaughter studied?

I made a mistake on the denominators in my first post, it should be 12 instead of 2. I tried to edit that, but it asked to put in the relevant formula. I can't find how to write formula here. I used to be able to write formula so I don't have to write on paper and scan it in. Seems like it changed since 10 years ago when I was active here.

- #7

yungman

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I am really not interested in this, I am drowning on my C++ programming already, just want to resolve this and help her on this one. I haven't even have a chance to work on my C++ today yet!

Thanks

- #8

yungman

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I have been looking at the question, the only values of t to get 2 binomials is this:

Is this right?

Is this right?

- #9

mfb

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Right.

- #10

SammyS

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I suppose that you are referring to the quadratic formula:I have to get hold of her. thanks

I made a mistake on the denominators in my first post, it should be 12 instead of 2. I tried to edit that, but it asked to put in the relevant formula. I can't find how to write formula here. I used to be able to write formula so I don't have to write on paper and scan it in. Seems like it changed since 10 years ago when I was active here.

The solution to the general quadratic equation: ##ax^2+bx+c=0 ##

is given by ##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

So, yes, you left the factor of ##a## out of the denominator. In this case ##a=6##.

In addition to what @Mark44 said, if ##b^2-4ac## is a perfect square, then ##ax^2+bx+c ## can be written as the product of two binomials. For your (granddaughter's) problem, you need ##t^2-600## to be a perfect square, (as well as being positive, of course).

This problem may be done this way, but it will be quite cumbersome. There are at least 6 possible values for t.

Another way to do this is to systematically find all of the pairs of binomials product give ##6x^2## for the leading term and ##25## for the constant term.

For example: ##(2x + ?)\times(3x + ?)## gives ##6x^2 + ... ##

and ##(?x + 1)\times(?x + 25)## gives ## ?x^2 +?x + 25##

- #11

- #12

yungman

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I use your logic to have this, but I have 4 variables A, B, C and D, and I cannot solve it.I suppose that you are referring to the quadratic formula:

The solution to the general quadratic equation: ##ax^2+bx+c=0 ##

is given by ##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

So, yes, you left the factor of ##a## out of the denominator. In this case ##a=6##.

In addition to what @Mark44 said, if ##b^2-4ac## is a perfect square, then ##ax^2+bx+c ## can be written as the product of two binomials. For your (granddaughter's) problem, you need ##t^2-600## to be a perfect square, (as well as being positive, of course).

This problem may be done this way, but it will be quite cumbersome. There are at least 6 possible values for t.

Another way to do this is to systematically find all of the pairs of binomials product give ##6x^2## for the leading term and ##25## for the constant term.

For example: ##(2x + ?)\times(3x + ?)## gives ##6x^2 + ... ##

and ##(?x + 1)\times(?x + 25)## gives ## ?x^2 +?x + 25##

##(Ax+B)(Cx+D)=6x^2 + tx + 25##

It would be nice to do this instead of using

##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

She have not learn this formula, it would be nice to have a way to do this without

##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

Last edited:

- #13

SammyS

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Looking at the other exercises in the image in the OP, I conclude that these exercises have to do with factoring trinomials into the product of binomials, all with integer coefficients.@SammyS: Why would you want it to be a perfect square? No one required rational roots or any other special requirement.

- #14

SammyS

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Correct. You have ##(Ax+B)(Cx+D)=6x^2 + tx + 25##.I use your logic to have this, but I have 4 variables A, B, C and D, and I cannot solve it.

##(Ax+B)(Cx+D)=6x^2 + tx + 25##

It would be nice to do this instead of using

##x = \dfrac{ -b \pm \sqrt{b^2-4ac} }{2a} ##

But A and C are related: ##A\cdot C = 6##. Also ##B\cdot D = 25##.

Now get systematic.

With A = 1 and C = 6, let B = 1 and D =25, and multiply ##(x+1)(6x+25) ## to get ##t##.

Then let B = 5 and D =5, and multiply ##(x+5)(6x+5) ## to get another value for ##t##.

Then let B = 25 and D = 1 ... . Don't forget, B and D can both be negative.

Now consider A = 2 and C = 3. Run through the cases for B and D again.

When can you stop?

- #15

yungman

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Ok, I expanded these out:Correct. You have ##(Ax+B)(Cx+D)=6x^2 + tx + 25##.

But A and C are related: ##A\cdot C = 6##. Also ##B\cdot D = 25##.

Now get systematic.

With A = 1 and C = 6, let B = 1 and D =25, and multiply ##(x+1)(6x+25) ## to get ##t##.

Then let B = 5 and D =5, and multiply ##(x+5)(6x+5) ## to get another value for ##t##.

Then let B = 25 and D = 1 ... . Don't forget, B and D can both be negative.

Now consider A = 2 and C = 3. Run through the cases for B and D again.

When can you stop?

I see the relation, I connected the pairs with same value of t. These are all the combination. But what does these means?

Last edited:

- #16

SammyS

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Yes. Also 53 .Ok, I expanded these out:

View attachment 270852

I see the relation, I connected the pairs with same value of t. These are all the combination. But what does these means?

Does this means t = 25, 31, 35, 77 and 151 all work?

Additionally, t can be the negative of each of these.

- #17

yungman

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Thanks, I missed the 53 when I read my notes. It's there also. I put it back in the last post already.Yes. Also 53 .

Additionally, t can be thenegative of each of these.

Thanks

Last edited:

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SammyS

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- #19

mfb

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I don't share that interpretation.Looking at the other exercises in the image in the OP, I conclude that these exercises have to do with factoring trinomials into the product of binomials, all with integer coefficients.

- #20

yungman

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Actually after sitting on it for a day, I really feel the answer in my

BUT, this is NOT the full set of answer. Remember, coef for both x and 25 can be

I still believe the answer in

Thanks

- #21

yungman

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Any comment?

- #22

Mark44

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Since there isn't any restriction given in the problem statement about solutions being integers or rational numbers, the solution in post #8 looks fine to me.Any comment?

However, I would simplify things a bit.

$$t \ge 10\sqrt 6 \text{ or } t \le -10\sqrt 6$$

If ##t = \pm 10\sqrt 6## then the binomial will have two repeated factors; i.e., it will have the form ##6(x - a)^2##.

- #23

yungman

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Thanks, I'll tell her to read this thread.Since there isn't any restriction given in the problem statement about solutions being integers or rational numbers, the solution in post #8 looks fine to me.

However, I would simplify things a bit.

$$t \ge 10\sqrt 6 \text{ or } t \le -10\sqrt 6$$

If ##t = \pm 10\sqrt 6## then the binomial will have two repeated factors; i.e., it will have the form ##6(x - a)^2##.

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