Solution to Quadratic Expression with Three Distinct Real Numbers

[anemone]

Here is this week's POTW:

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Fill in the blank:

If $a,\,b$ and $c$ are three distinct real numbers, then the quadratic expression \(\displaystyle \frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}\) is identically equal to _____.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. kaliprasad
2. castor28
3. MarkFL
4. Opalg
5. lfdahlSolution from castor28:

Spoiler

If $f(x)$ is the expression, we have $f(a)=f(b)=f(c)=1$, which means that the quadratic polynomial $f(x)-1$ has three roots. As a non-zero polynomial cannot have more roots than its degree, we have $f(x)-1=0$ and $f(x)=1$.

Solution from MarkFL:

Spoiler

Let's begin by writing:

\(\displaystyle \frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}=k\)

As $a,\,b,\,c$ are distinct, we may multiply through by $(a-b)(a-c)(b-c)$:

\(\displaystyle (x-b)(x-c)(b-c)-(x-c)(x-a)(a-c)+(x-a)(x-b)(a-b)=k(a-b)(a-c)(b-c)\)

Expand factors involving $x$ on the LHS:

\(\displaystyle (b-c)\left(x^2-(b+c)x+bc\right)-(a-c)\left(x^2-(a+c)x+ac\right)+(a-b)\left(x^2-(a+b)x+ab\right)=k(a-b)(a-c)(b-c)\)

Distribute and combine like terms:

\(\displaystyle a^2b-a^2c-ab^2+ac^2+b^2c-bc^2=k(a-b)(a-c)(b-c)\)

Factor the LHS:

\(\displaystyle (a-b)(a-c)(b-c)=k(a-b)(a-c)(b-c)\)

As $a,\,b,\,c$ are distinct, we may divide through by $(a-b)(a-c)(b-c)$ to obtain:

\(\displaystyle k=1\)

Thus, the given quadratic expression is identically equal to $1$.

The following member managed to find the intermediate but not the simplest result, so he gets partial credit:
1. topsquark