Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

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Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Let the normal number = ‘x’.

As per the inquiry,

We get the condition,

$x{}^\text{2}\text{ }\text{ }84\text{ }=\text{ }3\left( x+8 \right)$

$x{}^\text{2}\text{ }\text{ }84\text{ }=\text{ }3x\text{ }+\text{ }24$

$x{}^\text{2}\text{ }\text{ }3x\text{ }\text{ }84\text{ }\text{ }24\text{ }=\text{ }0$

$x{}^\text{2}\text{ }\text{ }3x\text{ }\text{ }108\text{ }=\text{ }0$

$x{}^\text{2}\text{ }\text{ }12x\text{ }+\text{ }9x\text{ }\text{ }108\text{ }=\text{ }0$

$x\left( x\text{ }\text{ }12 \right)\text{ }+\text{ }9\left( x\text{ }\text{ }12 \right)\text{ }=\text{ }0$

$\left( x\text{ }+\text{ }9 \right)\text{ }\left( x\text{ }\text{ }12 \right)$

⇒ x = – 9 and x = 12

Since, regular numbers can’t be negative.

The number is 12.

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