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Please help, I am stuck on the following question:

$$(x-a)^2=x^2-12x+b$$

Find the values of $a$ and $b$

Can you show me the working out?

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3  
Open term in LHS, compare both sides –  Awesome Mar 16 at 15:36

4 Answers 4

up vote 2 down vote accepted

If you don't know the theorem used implicitly in some of the other answers (that polynomial functions are identical iff they have equal coefficients), you can proceed more simply as follows. Expanding the square and subtracting it from the RHS yields

$$ (2a-12) x + b-a^2\, =\, 0$$

Evaluating the above at $\,x = 0\,$ yields $\,b-a^2 = 0\,$ so $\, b = a^2,\,$ so the above becomes

$$ (2a-12)x\, =\, 0$$

Evaluating the above at $\,x = 1\,$ yields $\,2a-12=0\,$ so $\, 2a=12,\,$ so $\, a=6,\,$ so $\, b = a^2 = 36.$

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Hints:

$$\begin{align*}(1)&\;\;(x-a)^2=x^2-12x+b\iff x^2-2ax+a^2=x^2-12x+b\\{}\\(2)&\;\;\text{Two polynomials are identical iff the coefficients of corresponding powers of $\;x\;$ are identical}\end{align*}$$

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Well, firstly you should expand the expression and get $$x^2-2ax+a^2=x^2-12x+b$$ then cross out $x^2$ $$a^2-2ax=b-12x$$ Which means that \begin{cases} -2ax = -12x \Rightarrow a=6\\ a^2=b \Rightarrow b=a^2=6^2=36\\ \end{cases} So $a=6$ and $b=36$

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The left-hand side equals $x^2-2ax+a^2$. Comparing linear coefficients, we get $-2a = -12$ or $a=6$. Comparing constant coefficients, we get $b=a^2=6^2=36$. Hence $a=6$ and $b=36$.

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