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Problem on Real numbers | Algebra | PRMO-2017 | Problem 18

Try this beautiful problem from Algebra based on real numbers from PRMO 2017. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra PRMO 2017 based on real numbers.

Problem on Real numbers | PRMO | Problem 18


If the real numbers \(x\), \(y\), \(z\) are such that \(x^2 + 4y^2 + 16z^2 = 48\) and \(xy + 4yz + 2zx = 24\). what is the
value of \(x^2 + y^2 + z^2\) ?

  • $24$
  • $21$
  • $34$

Key Concepts


Algebra

Equation

Check the Answer


Answer:\(21\)

PRMO-2017, Problem 18

Pre College Mathematics

Try with Hints


The given equation are

\(x^2 + 4y^2 + 16z^2 = 48\)
\(\Rightarrow (x)2 + (2y)2 + (4z)2 = 48\)
\(2xy + 8yz + 4zx = 48\)
adding tis equations we have to solve the problem….

Can you now finish the problem ……….

Now we can say that
\((x)^2 + (2y)^2 + (4z)^2 – (2xy) – (8yz) – (4zx) = 0\)
\(\Rightarrow [(x – 2y)2 + (2y – 4z)2 + (x – 4y)2)] = 0\)
\(x = 2y = 4z \)

\(\Rightarrow \frac{x}{4}=\frac{y}{2}=z\)

Can you finish the problem……..

Therefore we may say that,

\((x, y, z) = (4m, 2m, m)\)
\(x^2 + 4y^2 + 16z^2 = 48\)

\(16m^2 + 16m^2 + 16m^2 = 48\)
so \(m^2 = 1\)
\(x^2 + y^2 + z^2 = 21m^2 = 21\)

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