Pre-calc: Having trouble answering this question

[celticdove_]

Find real numbers b and c such that the following equation holds, where a is a real number:
1800a-10a^2=-10(a-b)^2 + c

 

[Dr.Peterson]

Find real numbers b and c such that the following equation holds, where a is a real number:
1800a-10a^2=-10(a-b)^2 + c

It's an odd question, which I think is probably meant to lead you toward (or else use your knowledge of) completing the square.

Presumably what it means is that they want you to make an expression in the form -10(a-b)^2 + c that is equivalent to 1800a-10a^2, in the sense that they will be equal for all values of a (not just for a specific given value of a).

It would make more sense if they had written it as

Find constant real numbers b and c such that the following equation holds for all x: 1800x-10x^2=-10(x-b)^2 + c​

To answer the question, you can just expand the right-hand side and match up coefficients: the coefficient of a^2 will be the same on both sides already; particular values of b and c will make the other two coefficients equal.

 

[celticdove_]

So if I expand the right, which is -10(x^2+2kx+k^2)+l, how would I go about matching up the coefficients?
Thanks for your response!

 

[JeffM]

I am assuming Dr. Peterson has correctly interpreted the question.

Where did the 1 come from, or is that a lower-case L?

[MATH]1800a - 10a^2 = - 10(a - b)^2 + c = - 10a^2 + 20ab - 10b^2 + c \ne -10(a^2 - 2ab + b^2) + 1.[/MATH]
Why the change of variables?

Equating coefficients. A quadratic with a variable in a has the form

[MATH]pa^2 + qa^1 + ra^0.[/MATH]
Three summands, each consisting of a power of a times a coefficient (which may be zero except for the coefficient on the term with a to the power of 2).

If we say that two quadratics in a are equal for all values of a

[MATH]pa^2 + qa^1 + ra^0 = ua^2 + va^1 + wa^0 \implies p = u,\ q = v, \text { and } r = w.[/MATH]
So for our specific problem,

[MATH]-10a^2 + 1800a^1 + 0 * a^0 = -10a^2 + 20ba^1 + (c - 20b^2)a^0 \implies -10 = -10, \ 20b = 1800, \text { and } c - 10b^2 = 0.[/MATH]

 

[lex]

Find constant real numbers [MATH]b[/MATH] and [MATH]c[/MATH] such that the following equation holds for all real numbers [MATH]x: 1800x-10x^2\equiv -10(x-b)^2 + c[/MATH]

Dr. Peterson's restatement of the problem makes it much easier to understand.
We want to find constants, [MATH]b[/MATH] and [MATH]c[/MATH]:
[MATH]-10x^2 + 1800x \equiv -10(x-b)^2+c \quad [/MATH] (1)

LHS = [MATH]-10x^2+1800x[/MATH][MATH]=-10(x^2-180x)[/MATH][MATH]=-10((x-90)^2-90^2)[/MATH][MATH]=-10(x-90)^2+10(90^2)[/MATH]and from this you can now read off the values of [MATH]b[/MATH] and [MATH]c[/MATH] in (1)