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Algebra Level 4

How many roots does the quadratic equation below, where a ≠ b ≠ c a \ne b \ne c are real, have?

( x + b ) ( x + c ) ( b − a ) ( c − a ) + ( x + c ) ( x + a ) ( c − b ) ( a − b ) + ( x + a ) ( x + b ) ( a − c ) ( b − c ) = 1 \large \dfrac{(x + b)(x + c)}{(b - a)(c - a)} + \dfrac{(x + c)(x + a)}{(c - b)(a - b)} + \dfrac{(x + a)(x + b)}{(a - c)(b - c)} = 1

Infinitely many roots Two real and distinct negative roots No real or complex root Two equal roots Two real and distinct positive roots

Ram Mohith by Ram Mohith , 18, India

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2 solutions

Brian Moehring
Oct 4, 2018

Set P ( x ) = ( x + b ) ( x + c ) ( b − a ) ( c − a ) + ( x + c ) ( x + a ) ( c − b ) ( a − b ) + ( x + a ) ( x + b ) ( a − c ) ( b − c ) − 1 P(x) = \frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} - 1

Then

  • since P ( x ) P(x) is the sum of three quadratics with real coefficients, we know that P ( x ) = A x 2 + B x + C P(x) = Ax^2 + Bx + C for real numbers A , B , C A,B,C .
  • P ( − a ) = ( − a + b ) ( − a + c ) ( b − a ) ( c − a ) + ( − a + c ) ( − a + c ) ( c − b ) ( a − b ) + ( − a + a ) ( − a + b ) ( a − c ) ( b − c ) − 1 = 1 + 0 + 0 − 1 = 0 \displaystyle P(-a) = \frac{(-a+b)(-a+c)}{(b-a)(c-a)} + \frac{(-a+c)(-a+c)}{(c-b)(a-b)} + \frac{(-a+a)(-a+b)}{(a-c)(b-c)} - 1 = 1 + 0 + 0 - 1 = 0
  • Similarly, P ( − b ) = P ( − c ) = 0 P(-b) = P(-c) = 0

Since a non-zero polynomial of degree ≤ 2 \leq 2 has at most 2 2 roots and we have shown that P ( x ) P(x) has at least three roots, it follows that P ( x ) = 0 P(x) = 0 identically. Therefore, ( x + b ) ( x + c ) ( b − a ) ( c − a ) + ( x + c ) ( x + a ) ( c − b ) ( a − b ) + ( x + a ) ( x + b ) ( a − c ) ( b − c ) = 1 \frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} = 1 for infinitely many \boxed{\text{infinitely many}} x x .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

( x + b ) ( x + c ) ( b − a ) ( c − a ) + ( x + c ) ( x + a ) ( c − b ) ( a − b ) + ( x + a ) ( x + b ) ( a − c ) ( b − c ) = 1 Multiply both sides by − ( a − b ) ( b − c ) ( c − a ) ( b − c ) ( x + b ) ( x + c ) + ( c − a ) ( x + c ) ( x + a ) + ( a − b ) ( x + a ) ( x + b ) = − ( a − b ) ( b − c ) ( c − a ) ( b − c ) ( x 2 + ( b + c ) x + b c ) + ( c − a ) ( x 2 + ( c + a ) x + c a ) + ( a − b ) ( x 2 + ( a + b ) x + a b ) = a 2 b − a b 2 + b 2 c − b c 2 + c 2 a − c a 2 ( b − c + c − a + a − b ) x 2 + ( b 2 − c 2 + c 2 − a 2 + a 2 − b 2 ) x + b c ( b − c ) + c a ( c − a ) + a b ( a − b ) = a 2 b − a b 2 + b 2 c − b c 2 + c 2 a − c a 2 b c ( b − c ) + c a ( c − a ) + a b ( a − b ) = a 2 b − a b 2 + b 2 c − b c 2 + c 2 a − c a 2 a 2 b − a b 2 + b 2 c − b c 2 + c 2 a − c a 2 ≡ a 2 b − a b 2 + b 2 c − b c 2 + c 2 a − c a 2 \begin{aligned} \frac {(x+b)(x+c)}{(b-a)(c-a)} + \frac {(x+c)(x+a)}{(c-b)(a-b)} + \frac {(x+a)(x+b)}{(a-c)(b-c)} & = 1 \quad \quad \small \color{#3D99F6} \text{Multiply both sides by }-(a-b)(b-c)(c-a) \\ (b-c)(x+b)(x+c) + (c-a)(x+c)(x+a) + (a-b)(x+a)(x+b) & = -(a-b)(b-c)(c-a) \\ (b-c)(x^2+(b+c)x+bc) + (c-a)(x^2+(c+a)x+ca) + (a-b)(x^2+(a+b)x+ab) & = a^2b-ab^2 + b^2c - bc^2 + c^2a - ca^2 \\ (b-c+c-a+a-b)x^2+(b^2-c^2+c^2-a^2+a^2-b^2)x + bc(b-c) + ca(c-a)+ab(a-b) & = a^2b-ab^2 + b^2c - bc^2 + c^2a - ca^2 \\ bc(b-c) + ca(c-a)+ab(a-b) & = a^2b-ab^2 + b^2c - bc^2 + c^2a - ca^2 \\ a^2b-ab^2 + b^2c - bc^2 + c^2a - ca^2 & \equiv a^2b-ab^2 + b^2c - bc^2 + c^2a - ca^2 \end{aligned}

Note that the LHS is identical to the RHS. The equation is independent of x x . It is true for all value of x x . Therefore, it has infinitely many roots .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

@Ram Mohith You have to mention a a , b b and c c are real and they are not equal to each other. Otherwise the equation is undefined. We don't need a space before a question mark (?), comma (,) or period (.).

Chew-Seong Cheong - 2 years, 8 months ago

I have a still easier method :

  • Put x = a x = a in the equation and you will get it as 1 = 1 1 = 1 . So, x = a x = a is a root.
  • Put x = b x = b in the equation and you will get it as 1 = 1 1 = 1 . So, x = b x = b is also a root.
  • Put x = c x = c in the equation and you will get it as 1 = 1 1 = 1 . So, x = c x = c is also a root.

As a , b , c a,b,c are any real numbers and are also the roots of given equation there are infinite roots.

Ram Mohith - 2 years, 8 months ago

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