If x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2, then the difference between z and x is:
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 5
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Concept |
Understanding |
Application |
Fair calculation |
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(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz
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Given x + y + z = 19 x2 + y2 + z2 = 133 xz = y2 |
361 = 133 + 2(xy + yz + xz) Put xz = y2 ⇒ 361 = 133 + 2(xy + yz + y2) ⇒ 361 = 133 + 2y (x + y + z) ⇒ y (x + y + z) = 114 ⇒ y = 6 And xz = 36 Since, x + y + z = 19 ⇒ x + z = 19 – 6 = 13 |
336 – 133 = 228 228/2 = 114 |
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(x + z)2 = x2 + z2 + 2x |
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Now (x + z)2 = x2 + z2 + 2xz x2 + z2 = 169 – 72 = 97 (x - z)2 = 97 – 72 = 25 x – z = 5
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√25 = 5 |