Expanding brackets: find a, q for (3x - a)^2 = 9x^2 + qx + 4

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Jake101

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I get a) but b) is beyond me.


10. (a) Expand (3x - a)2.

. . ..(b) Hence find the value of a and q for which the following is true:

. . . . . . .\(\displaystyle (3x\, -\, a)^2\, =\, 9x^2\, +\, qx\, +\, 4\)


I could do it 20 years ago but the years have eroded my beain cells. The answer is:

a = 2, q = -12

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Jake101 said:
I get a) but b) is beyond me.


10. (a) Expand \(\displaystyle (3x\, -\, a)^2.\)

. . ..(b) Hence find the value of a and q for which the following is true:

. . . . . . .\(\displaystyle (3x\, -\, a)^2\, =\, 9x^2\, +\, qx\, +\, 4\)


I could do it 20 years ago but the years have eroded my beain cells. The answer is:

a = 2, q = -12
Click to expand...
What expression did you get for (a)?
 
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Jake101 said:
9x^2 - 6xa + a^2
Click to expand...
That's correct.

Did you notice that the left-hand side of the equation in part (b) is the same expression as given in part (a)?

In other words, in part (b), replace the left-hand side with:

9x^2 - 6ax + a^2

(I rewrote the coefficient of your middle term, to show that x is multiplied by 6a.)

Now, compare the two sides of the equation. Does this give you any ideas, for finding values for a and q?
 
Since x^2 = 4, x = 2.
Thats as far as i go.

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Jake101 said:
Since x^2 = 4, x = 2.
Click to expand...
I'm thinking that you intended to write:

Since a^2 = 4, a = 2.


That's one possibility. Here's another:

a = -2


After expanding the left-hand side of the equation in part (b), we have:

9x^2 - 6ax + a^2 = 9x^2 + qx + 4

Comparing the coefficients of x-terms on each side, we see that -6a must equal q.

You can determine the value of q, in each case.

Case I: a = 2

Case II: a = -2

There are two answers, to this exercise, because there are two cases to consider. :cool:
 
Jake101 said:
II get a) but b) is beyond me.


10. (a) Expand (3x - a)2.

. . ..(b) Hence find the value of a and q for which the following is true:

. . . . . . .\(\displaystyle (3x\, -\, a)^2\, =\, 9x^2\, +\, qx\, +\, 4\)


I could do it 20 years ago but the years have eroded my beain cells. The answer is:

a = 2, q = -12
Click to expand...
Then you get:

9x^2 - 6a * x + a^2 = 9x^2 + q * x + 4

 - 6a * x + a^2 = q * x + 2^2

Now equate the constant terms - 'a^2' and '2^2'.

What does it tell you about the value of 'a'?
 
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Subhotosh Khan said:
Then you get:

9x^2 - 6a * x + a^2 = 9x^2 + q * x + 4

 - 6a * x + a^2 = q * x + 2^2

Now equate the constant terms - 'a^2' and '2^2'.

What does it tell you about the value of 'a'?
Click to expand...
The value of x is 2 or -2

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Jake101 said:
Sorry i mean the value of a is 2 or -2

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Click to expand...
I think i got it.
a39cd3cb8df6b4c5dfe827e5465dd9b8.jpg


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mmm4444bot said:
That's correct.

Did you notice that the left-hand side of the equation in part (b) is the same expression as given in part (a)?

In other words, in part (b), replace the left-hand side with:

9x^2 - 6ax + a^2

(I rewrote the coefficient of your middle term, to show that x is multiplied by 6a.)

Now, compare the two sides of the equation. Does this give you any ideas, for finding values for a and q?
Click to expand...
Yes. Ive got it. Thanks.


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