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

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 \(x^2 + y^2 + z^2\) ?

- $24$
- $21$
- $34$

Algebra

Equation

But try the problem first...

Answer:\(21\)

Source

Suggested Reading

PRMO-2017, Problem 18

Pre College Mathematics

First hint

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 ..........

Second Hint

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........

Final Step

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 \(m^2 = 1\)

\(x^2 + y^2 + z^2 = 21m^2 = 21\)

- https://www.cheenta.com/quadratic-equation-isi-b-stat-objective-problem-198/
- https://www.youtube.com/watch?v=pYSIvF7jZy4

Contents

[hide]

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

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 \(x^2 + y^2 + z^2\) ?

- $24$
- $21$
- $34$

Algebra

Equation

But try the problem first...

Answer:\(21\)

Source

Suggested Reading

PRMO-2017, Problem 18

Pre College Mathematics

First hint

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 ..........

Second Hint

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........

Final Step

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 \(m^2 = 1\)

\(x^2 + y^2 + z^2 = 21m^2 = 21\)

- https://www.cheenta.com/quadratic-equation-isi-b-stat-objective-problem-198/
- https://www.youtube.com/watch?v=pYSIvF7jZy4

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