Multiple choice

If the roots of the equation $\left( x-a \right) \left( x-b \right) +\left( x-b \right) \left( x-c \right) +\left( x-c \right) \left( x-a \right) =0$ are equal, then ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$ is equal to

  1. $a+b+c$
  2. $2a+b+c$
  3. $3abc$
  4. $ab+bc+ca$
  5. $abc$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Expanding the equation gives 3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0. For equal roots, the discriminant D = 0. So, 4(a+b+c)^2 - 4*3*(ab+bc+ca) = 0, which simplifies to a^2+b^2+c^2 - ab - bc - ca = 0. This implies a=b=c, meaning a^2+b^2+c^2 = ab+bc+ca.

AI explanation

Expanding the given equation gives (x^2 - ax - bx + ab) + (x^2 - bx - cx + bc) + (x^2 - cx - ax + ac) = 0. Combining like terms results in 3x^2 - 2(a + b + c)x + (ab + bc + ca) = 0. For this quadratic equation to have equal roots, its discriminant must be zero, so (-2(a + b + c))^2 - 4(3)(ab + bc + ca) = 0. Expanding the square gives 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 12ab - 12bc - 12ca = 0. Dividing by 4 simplifies this to a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca = 0, which reduces to a^2 + b^2 + c^2 - ab - bc - ca = 0. Therefore, a^2 + b^2 + c^2 = ab + bc + ca.

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