The sum of two numbers is 9 and the sum of their squares is 41 . Taking one number as x, form an equation in x and solve it to find the numbers .

Taking first number as x

Since the sum of two numbers is 9 , so other number is 9 - x

Given the sum of squares of numbers = 41

$\Rightarrow x^2 + (9 - x)^2 = 41 \\[1em] \Rightarrow x^2 + x^2 + 81 - 18x = 41 \\[1em] \Rightarrow 2x^2 - 18x + 81 - 41 = 0 \\[1em] \Rightarrow 2x^2 - 18x + 40 = 0 \\[1em] \Rightarrow 2(x^2 - 9x + 20) = 0 \\[1em] \Rightarrow x^2 - 9x + 20 = 0 \\[1em] \Rightarrow x^2 - 5x - 4x + 20 = 0 \\[1em] \Rightarrow x(x - 5) - 4(x - 5) = 0 \\[1em] \Rightarrow (x - 5)(x - 4) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 5 \text{ or } x = 4$

∴ x = 5 , 9 - x = 4

Equation in x = x² + (9 - x)² = 41 ; Hence, required numbers are 4 and 5.