Question

the sum of the squares of two consecutive natural numbers is 421.Find the number​

Answer 1

Let one of the numbers be x

So, other number = x + 1

It is given that,

x^2 + (x+1)^2 = 421

=> x^2 + x^2 + 1 + 2x = 421

=> 2x^2 + 2x - 420 = 0

=> x^2 + x - 210 = 0

=> x^2 + 15x - 14x - 210 = 0

=> x(x +15) - 14(x + 15) = 0

=> (x+15)(x-14) = 0

So, either x = -15 or x = 14

Since it is given that x is a natural number, hence rejecting x = -15

So, x = 14

Thus the two numbers are 14 and (14 + 1) i.e., 15

Follow me please.

Answer 2

$\sf\huge\blue{\underline{\underline{ Question : }}}$

The sum of the squares of two consecutive natural numbers is 421.Find the number.

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• Sum of two consecutive natural numbers is 421.

★ To find,

• The two numbers.

★ Let,

$\tt\:\implies The\:first\:natural\:number = x$

$\tt\:\implies The\:second\:natural\:number= (x + 1).....(1)$

★ According to the question,

➡ Sum of squares of the natural numbers = 421

$\bf\:\implies x^{2} + (x + 1)^{2} = 421$

• (a + b)² = a² + b² + 2ab

$\bf\:\implies x^{2} + (x)^{2} + (1)^{2} + 2(x)(1) = 421$

$\bf\:\implies x^{2} + x^{2} + 1 + 2x = 421$

$\bf\:\implies 2x^{2} + 2x + 1 - 421 = 0$

$\bf\:\implies 2x^{2} + 2x - 420 = 0$

$\bf\:\implies 2(x^{2} + x - 210) = 0$

$\bf\:\implies x^{2} + x - 210 = \frac{0}{2}$

$\bf\:\implies x^{2} + x - 210 = 0$

$\bf\:\implies x^{2} + 15x - 14x - 210 = 0$

$\bf\:\implies x(x + 15) - 14(x + 15) = 0$

$\bf\:\implies (x + 15)(x - 14) = 0$

$\bf\:\implies x = - 15 \: ; \: 14$

• [ Since, decline the negative integers. ]

$\bf\:\implies x = 14$

• Substitute x in (1).

$\bf\:\implies 14 + 1$

$\bf\:\implies 15$

$\underline{\boxed{\rm{\purple{ \therefore The\:two\:natural\:numbers\:are\: 14 \: and \: 15.}}}}\:\orange{\bigstar}$

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