You visited us 1 times! Enjoying our articles? Unlock Full Access!

Derivative from First Principle

If x - 12 +...

Question

If $(x - 1)^{2} + (y - 3)^{2} + (z - 5)^{2} + (t - 7)^{2} = 0$ then $x y z t + 16$ is equal to

5^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

12^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $11^{2}$
Power of every factor is even, so to hold given equation individual factor has to be zero
Hence
$x - 1 = 0$
$y - 3 = 0$
$z - 5 = 0$
$t - 7 = 0$
Hence
$x = 1, y = 3, z = 5, t = 7$
Therefore
$x y z t + 16$
$= (3.5.7) + 16$
$= 105 + 16$
$= 121$
$= 11^{2}$
Hence, option 'C' is correct.

Suggest Corrections

Similar questions

$1^{2} + 3^{2} + 5^{2} + 7^{2} + 9^{2}$ =___.

1^{2} + 2^{2} - 3^{2} + 4^{2} + 5^{2} - 6^{2} + 7^{2} + 8^{2} - 9^{2} + . . . .

3 n

1^{2} = \frac{1}{6} [1 \times (1 + 1) \times (2 \times 1 + 1)] 1^{2} + 2^{2} = \frac{1}{6} [2 \times (2 + 1) \times (2 \times 2 + 1)] 1^{2} + 2^{2} + 3^{2} = \frac{1}{6} [3 \times (3 + 1) \times (2 \times 3 + 1)] 1^{2} + 2^{2} + 3^{2} + 4^{2} = \frac{1}{6} [4 \times (4 + 1) \times (2 \times 4 + 1)]

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + . . . . . . = 21^{2}

5^{2} + . . . . . + 30^{2} = 31^{2}

6^{2} + 7^{2} + . . . . . . = . . . . . . . .

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} +

= 21^{2}

5^{2} +

+ 30^{2} = 31^{2}

6^{2} + 7^{2} +

=

Join BYJU'S Learning Program