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1987 Paper 2 Q1
D: 1500.0 B: 1500.0
quadratics algebraic identity proof inequality generalisation expanding brackets sum of squares

Prove that:

  1. if \(a+2b+3c=7x\), then \[ a^{2}+b^{2}+c^{2}=\left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2}; \]
  2. if \(2a+3b+3c=11x\), then \[ a^{2}+b^{2}+c^{2}=\left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2}. \]
Give a general result of which \((i) \)and \((ii) \)are special cases.

Solution:

  1. \begin{align*} \left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\ &= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\ &= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\ &= a^2 + b^2 + c^2 \end{align*}
  2. \begin{align*} \left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\ &= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\ &= a^2+b^2+c^2 \end{align*}
The general result is: If \(\frac{A^2+B^2+C^2}{2}x =Aa+Bb+Cc\) then \((Ax-a)^2 + (Bx-b)^2 + (Cx-c)^2 = a^2+b^2+c^2\) Alternatively, if \(\lambda = \frac{2\mathbf{x} \cdot \mathbf{y}}{\Vert x \Vert^2}\) then \(\Vert \lambda \mathbf{x} - \mathbf{y}\Vert^2 = \Vert \mathbf{y} \Vert^2\) which is easy to see is true.

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