If x + y + z = 19, x2 +y2 + Z2 = 133 and xz = y2, x > z > 0, what is the value of (x - z)?
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Correct answer: B
Explanation
The correct answer is Option 2 i.e 5.
Given:
x + y + z = 19
x2 + y2 + z2 = 133
xz = y2
Formula Used:
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
Calculation:
⇒ x + z = 19 - y
⇒ x2 + z2 = 133 - y2
Square on both sides
⇒ x2 + z2 + 2xz = 361 + y2 - 38y
⇒ 133 - y2 + 2y2 = 361 + y2 - 38y [x2 + z2 = 133 - y2 and xz = y2]
⇒ 38y = 361 - 133 = 228
⇒ y = 6
⇒ (x - z)2 = x2 + z2 - 2xz
⇒ (x - z)2 = 133 - y2 - 2y2
⇒ (x -z)2 = 133 - 3y2
⇒ (x - z)2 = 133 - 3 × 62
⇒ (x - z)2 = 25
⇒ (x - z) = √25 = 5
The difference between z and x is 5