Question : If $x + y + z = 19, x^2 + y^2 + z^2 = 133$ and $xz = y^2, x > z > 0,$ what is the value of $(x - z)$?
Option 1: 5
Option 2: 0
Option 3: –2
Option 4: –5
Correct Answer: 5
Solution : Given: $x^2 + y^2 + z^2 = 133⇒ x^2 + z^2 = 133 - y^2$ $xz = y^2$ $x + y + z = 19$ $⇒ x + z = 19 - y$ Squaring both sides, we get, $⇒ x^2 + z^2 + 2xz = 361 + y^2 - 38y $ $⇒ 133 - y^2 + 2y^2 = 361 + y^2- 38y$ $⇒ 38y = 361 -133 = 228$ $\therefore y = 6$ Now, $(x - z)^2 = x^2 + z^2 - 2xz$ $⇒(x - z)^2 = 133 - y^2 – 2y^2$ $⇒(x - z)^2 = 133 - 3y^2$ $⇒(x - z)^2 = 133 - 3 × 6^2$ $⇒(x - z)^2 = 25$ $⇒(x - z) = \sqrt{25} = 5$ Hence, the correct answer is 5.
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