Question

for real numbers , x ,y and z if 9x²+16y²+4z²=108 and 6xy+4yz+3zx= 54 then find the value of x²+y²+z²​

Answer 1

Answer:

Let 3x = a , 4y = b and 2z=c . Our equations become

${a}^{2} + {b}^{2} + {c}^{2} = 108 \\ ab + bc + ca = 2 \times 54 = 108 \\ so \\ {a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca = 108 \\now \\ \\ {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca =0 \\ \\ \frac{ {(a - b)}^{2} + (b - c) {}^{2} + ( {c - a)}^{2} }{2} = 0 \\ \\ which \: \: is \: \: only \: \: possible \: \: if \: \: \\ (a - b) {}^{2} = (b - c) {}^{2} =( c - a) {}^{2} \\ or \\ a = b = c \\ so \: \: we \: \: get \\ \\ {a}^{2} + {b}^{2} + {c}^{2} = 108 \\ \\ {a}^{2} = {b}^{2} = {c}^{2} = \frac{108}{3} = 36 \\ \\ so \\ \\ {x}^{2} + {y}^{2} + {z}^{2} = {a}^{2} ( \frac{1}{9} + \frac{1}{16} + \frac{1}{4} ) = \frac{61}{4}$