Binomial theorem: How did they get this answer?

[draco134]

Hi, I'm a little new here and i need a little help. I have a question on the TLE 12 program that even my teacher doesnt understand. Both he and i dont know how they get the answer, can you guys help out. It would really help me be able to complete my course.

In the expansion of (ax + b)[sup:20f7emtd]4[/sup:20f7emtd], the coefficient of the second term is 1500 and the coefficient of the third term is 1350. Given that a and b are both positive, find the values of a and b.

Apply the Binomial Theorem to obtain expressions for the second and third terms.

. . .t[sub:20f7emtd]2[/sub:20f7emtd] = [sub:20f7emtd]4[/sub:20f7emtd]C[sub:20f7emtd]1[/sub:20f7emtd](ax)[sup:20f7emtd]3[/sup:20f7emtd](b)[sup:20f7emtd]1[/sup:20f7emtd]

. . . . .= 4a[sup:20f7emtd]3[/sup:20f7emtd]bx[sup:20f7emtd]3[/sup:20f7emtd]

. . .t[sub:20f7emtd]3[/sub:20f7emtd] = [sub:20f7emtd]4[/sub:20f7emtd]C[sub:20f7emtd]2[/sub:20f7emtd](ax)[sup:20f7emtd]2[/sup:20f7emtd](b)[sup:20f7emtd]2[/sup:20f7emtd]

. . . . .= 6a[sup:20f7emtd]2[/sup:20f7emtd]b[sup:20f7emtd]2[/sup:20f7emtd]x[sup:20f7emtd]2[/sup:20f7emtd]

Therefore, 4a[sup:20f7emtd]3[/sup:20f7emtd]b = 1500 (1) and 6a[sup:20f7emtd]2[/sup:20f7emtd]b[sup:20f7emtd]2[/sup:20f7emtd] = 1350 (2).

From (1):

. . . . .4a[sup:20f7emtd]3[/sup:20f7emtd]b = 1500

. . . . .b = 375 / a[sup:20f7emtd]3[/sup:20f7emtd]

Substitute into (2) using the above:

. . . . .6a[sup:20f7emtd]2[/sup:20f7emtd]b[sup:20f7emtd]2[/sup:20f7emtd] = 1350

. . . . .6a[sup:20f7emtd]2[/sup:20f7emtd] (375 / a[sup:20f7emtd]3[/sup:20f7emtd])[sup:20f7emtd]2[/sup:20f7emtd] = 1350

. . . . .375[sup:20f7emtd]2[/sup:20f7emtd] / a[sup:20f7emtd]4[/sup:20f7emtd] = 225

. . . . .a[sup:20f7emtd]4[/sup:20f7emtd] = 375[sup:20f7emtd]2[/sup:20f7emtd] / 225

. . . . .a[sup:20f7emtd]4[/sup:20f7emtd] = 625

. . . . .a = +/- 5

Since a is positive, then a = 5. Substitute a = 5 into b = 375 / a[sup:20f7emtd]3[/sup:20f7emtd]:

. . .b = 375 / a[sup:20f7emtd]3[/sup:20f7emtd]

. . . . .= 375 / 5[sup:20f7emtd]3[/sup:20f7emtd]

. . . . .= 3

Therefore a = 5 and b = 3.

these are questions related to the Binomial Theorem.
Thanks
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edited by stapel -- Reason for edit: replacing unreadable graphics of text with legible text

 

[Deleted member 4993]

Re: Binomial theorem problem

Hi, I'm a little new here and i need a little help. I have a question on the TLE 12 program that even my teacher doesnt understand. Both he and i dont know how they get the answer, can you guys help out. It would really help me be able to complete my course.

Heres a screen print of the problem


and heres a screen print of the answer of the problem


and heres the other part


these are questions related to the Binomial Theorem.
Thanks

First write the expanded form of (ax + b)^4

Then you will get two equations from the given two conditions - in terms of 'a' and 'b'.

Then solve for 'a' and 'b'.

The answer shown in your book is very complete (following the method I wrote above) - what part of that you don't understand? (by the way, if your teacher does not undertand this - find a new teacher - now)

If you need more help - write back showing your work and indicate exactly where you are stuck.

 

[pka]

Re: Binomial theorem problem

The real difficulty with this sort of a problem is: What you may call the second term, I may call the same term the fourth term. And we would both be correct. It all depends upon the method of expansion.
\(\displaystyle \left( {ax + b} \right)^4 = \sum\limits_{k = 0}^4 {{ 4 \choose k }\left( {ax} \right)^k b^{4 - k} } = \sum\limits_{k = 0}^4 {{ 4 \choose k }\left( {ax} \right)^{4 - k} b^k }\).
Both of those are correct expansions of your expression. But the terms do not appear in the same order. If we stipulate that the powers of x must decrease, then the second summation is what you want.
\(\displaystyle \left( {ax + b} \right)^4 = \sum\limits_{k = 0}^4 {{ 4 \choose k }\left( {ax} \right)^{4 - k} b^k } = a^4 x^4 + 4a^3 bx^3 + 6a^2 b^2 x^2 + 4ab^3 x + b^4\).

 

[draco134]

Re: Binomial theorem problem

ok ive gotten this written out

4 a^3 b x^3=1500
and
6a^2 b^2 x^2=1350

and whats now getting me is that the (x) dissappears from both equations so now they look like

4 a^3 b =1500
and
6a^2 b^2 =1350

Why does the (X) variable disappear

 

[stapel]

Why does the (X) variable disappear

If you're working with the coefficients of the variable, why would you include the variable? :wink:

Eliz.

 

[Deleted member 4993]

In the expansion of (ax + b)[sup:23ixvupt]4[/sup:23ixvupt], the coefficient of the second term is 1500 and the coefficient of the third term is 1350. Given that a and b are both positive, find the values of a and b.

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