((x-a)(x-b))/((c-a)(c-b)) + ((x-b)(x-c))/((a-b)(a-c)) + ((x-a)(x-c))/((b-a)(b-c)) = 1

[nasa]

Hello, I can not do this exercise:

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Given the following equation:

. . . . .$\displaystyle \dfrac{(x\, -\, a)(x\, -\, b)}{(c\, -\, a)(c\, -\, b)}\, +\, \dfrac{(x\, -\, b)(x\, -\, c)}{(a\, -\, b)(a\, -\, c)}\, +\, \dfrac{(x\, -\, a)(x\, -\, c)}{(b\, -\, a)(b\, -\, c)}\, =\, 1$

The constants a, b, and c are distinct real numbers.

i. Confirm that a, b, and c are solutions to the equation.
ii. Can we say that this equation is of the second degree?

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:shock:I managed to find for when a, b and c are solutions to equation (I simply replace x by successively a, b and c). But I have to develope to find if it is a quadratic equation?

 

[stapel]

Given the following equation:

. . . . .$\displaystyle \dfrac{(x\, -\, a)(x\, -\, b)}{(c\, -\, a)(c\, -\, b)}\, +\, \dfrac{(x\, -\, b)(x\, -\, c)}{(a\, -\, b)(a\, -\, c)}\, +\, \dfrac{(x\, -\, a)(x\, -\, c)}{(b\, -\, a)(b\, -\, c)}\, =\, 1$

The constants a, b, and c are distinct real numbers.

i. Confirm that a, b, and c are solutions to the equation.
ii. Can we say that this equation is of the second degree?

---

:shock:I managed to find for when a, b and c are solutions to equation (I simply replace x by successively a, b and c). But I have to develope to find if it is a quadratic equation?

How does your book define a "second-degree" polynomial? If you multiply out this polynomial, what will be the leading degree? How does this compare with your book's definition?

Please be complete. Thank you!

 

[pka]

Given the following equation:
. . . . .$\displaystyle \dfrac{(x\, -\, a)(x\, -\, b)}{(c\, -\, a)(c\, -\, b)}\, +\, \dfrac{(x\, -\, b)(x\, -\, c)}{(a\, -\, b)(a\, -\, c)}\, +\, \dfrac{(x\, -\, a)(x\, -\, c)}{(b\, -\, a)(b\, -\, c)}\, =\, 1$
The constants a, b, and c are distinct real numbers.
i. Confirm that a, b, and c are solutions to the equation.
ii. Can we say that this equation is of the second degree?

Rewrite it as:
.$\displaystyle \dfrac{(x-a)(x-b)}{(a-c)(b-c)}\, +\, \dfrac{(x- b)(x- c)}{(a - b)(a - c)}\, -\, \dfrac{(x- a)(x - c)}{(a-b)(b-c)}\, =\, 1$

Use direct substitution.

 

[Ishuda]

Just as an aside, note that an second order interpolating polynomial p(x) for the three points (a, f(a)), (b, f(b)) and (c, f(c)) is given by
.$\displaystyle p(x)\, =\, f(c)\, \dfrac{(x\, -\, a)(x\, -\, b)}{(c\, -\, a)(c\, -\, b)}\, +\, f(a)\, \dfrac{(x\, -\, b)(x\, -\, c)}{(a\, -\, b)(a\, -\, c)}\, +\, f(b)\, \dfrac{(x\, -\, a)(x\, -\, c)}{(b\, -\, a)(b\, -\, c)}$

In this case since f(a)=f(b)=f(c)=1 we must have a, b, c are solutions to p(x)=1.

EDIT: to follow up on the gentle hint provided above: Note that a, b, and c must be distinct (otherwise there would be a division by zero). Now, if p(x) [the given expression with f(a)=f(b)=f(c)=1)] were a quadratic, then p(x)-1 would also be a quadratic and have three solutions. Since a quadratic=0 has at most two distinct solutions, p(x)-1 is not a quadratic and thus neither is p(x). By inspection, x² is the greatest power of x for the given expression, thus it can not be something higher than second order. That leaves linear (first order) and constant (zeroth order). But a first order has at most only one zero, thus it is not first order. That leaves only that the expression is constant.

 

[MGM]

Nope.Equation is not of second degree.

Substitution is a good method, but I just went totally brute-force here, and found that the coefficients of x^2 is 0.

 

[Deleted member 4993]

Hello, I can not do this exercise:

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Given the following equation:

. . . . .$\displaystyle \dfrac{(x\, -\, a)(x\, -\, b)}{(c\, -\, a)(c\, -\, b)}\, +\, \dfrac{(x\, -\, b)(x\, -\, c)}{(a\, -\, b)(a\, -\, c)}\, +\, \dfrac{(x\, -\, a)(x\, -\, c)}{(b\, -\, a)(b\, -\, c)}\, =\, 1$

The constants a, b, and c are distinct real numbers.

i. Confirm that a, b, and c are solutions to the equation.
ii. Can we say that this equation is of the second degree?

---

:shock:I managed to find for when a, b and c are solutions to equation (I simply replace x by successively a, b and c). But I have to develope to find if it is a quadratic equation?

$\displaystyle \dfrac{(x\, -\, a)(x\, -\, b)}{(c\, -\, a)(c\, -\, b)}\, +\, \dfrac{(x\, -\, b)(x\, -\, c)}{(a\, -\, b)(a\, -\, c)}\, +\, \dfrac{(x\, -\, a)(x\, -\, c)}{(b\, -\, a)(b\, -\, c)}\, =\, 1$

$\displaystyle -\dfrac{(x\, -\, a)(x\, -\, b)(a\, -\, b)}{(c\, -\, a)(b\, -\, c)(a\, -\, b)}\, - \, \dfrac{(x\, -\, b)(x\, -\, c)(b\, -\, c)}{(a\, -\, b)(c\, -\, a)(b\, -\, c)}\, - \, \dfrac{(x\, -\, a)(x\, -\, c)(c\, -\, a)}{(a\, -\, b)(b\, -\, c)(c\, -\, a)}\, =\, 1$

$\displaystyle \dfrac{x^2(a\, -\, b) - x (a^2 - b^2) + ab(a-b) \ + \ x^2(b\, -\, c) - x (b^2 - c^2) + bc(b - c) \ + \ x^2(c\, -\, a) - x (c^2 - a^2) + ca(c - a)}{(a\, -\, b)(b\, -\, c)(c\, -\, a)}\, =\, -1$

$\displaystyle \dfrac{ab(a-b) \ + bc(b - c) \ + \ ca(c - a)}{(a\, -\, b)(b\, -\, c)(c\, -\, a)}\, =\, -1$

Thus the solution for 'x' is any entity - because the given equation reduces to being independent of 'x'.