Question

the sum of the squares pf two natural numbers is 29. if second number is one more than twice the first number . find the number​

Answer 1

Let the two numbers be x and y.

Then,

$\sf\red{{x}^{2} + {y}^{2} = 29 \: and \: y = 2x + 1}$

$\therefore$ $\sf\red{ {x}^{2} + {(2x + 1)}^{2} = 29}$

$\longrightarrow$$\sf\red{{5x}^{2} + 4x - 28 = 0}$

$\longrightarrow$$\sf{(x-2)(5x+14)=0}$

$\longrightarrow$$\sf{x=2\:or\:x=- \frac{14}{5}}$

As x is a natural number, $\sf{x ≠ -\frac{14}{5}}$

So, x = 2,y = 2 × 2 + 1 = 5

Hence, the two natural numbers are 2 and 5.