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Solving a Quadratic Equation by Completion of Squares Method

If a2 + b2 ...

Question

If $(a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) = (a x + b y + c z)^{2};$ show that $x : a = y : b = z : c .$

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Solution

$(a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) = {(a x + b y + c z)}^{2} a^{2} x^{2} + a^{2} y^{2} + a^{2} z^{2} + b^{2} x^{2} + b^{2} y^{2} + b^{2} z^{2} + c^{2} x^{2} + c^{2} y^{2} + c^{2} z^{2} = a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} + 2 a b x y + 2 b c y z + 2 a c z x$
Rearranging the terms
$b^{2} x^{2} + a^{2} y^{2} - 2 a b x y + c^{2} y^{2} + b^{2} z^{2} - 2 b c y z + c^{2} x^{2} + a^{2} z^{2} - 2 a c z x = 0 {(b x - a y)}^{2} + {(c y - b z)}^{2} + {(c x - a z)}^{2} = 0 \Rightarrow b x - a y = 0, c y - b z = 0, c x - a z = 0 \Rightarrow \frac{x}{a} = \frac{y}{b}, \frac{y}{b} = \frac{z}{c}, \frac{z}{c} = \frac{x}{a} \Rightarrow \frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

hence proved.

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