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

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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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

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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