Respuesta :
The numbers are 45, 54
Solution:
Given a 2 digit number:
a = the 10's digit of the 1st number
b = units digit of the 1st number
From given information,
The sum of both digits, of either of two two-digit numbers, in whatever order the digits are written, is 9
a + b = 9 ---- eqn 1
The square of either of the digits of either number, minus the product of both digits, plus the square of the other digit is the number 21
Square of "a" - product of "a" and "b" + square of "b" = 21
[tex]a^2 - ab + b^2 = 21[/tex] --- eqn 2
Let us solve eqn 1 and eqn 2
From eqn 1,
a = 9 - b --- eqn 3
Substitute eqn 3 in eqn 2
[tex](b - 9)^2 - (9 - b)(b) + b^2 = 21\\\\b^2 + 81 - 18b - 9b + b^2 + b^2 = 21\\\\3b^2 - 27b + 60 = 0[/tex]
Let us solve the above equation using quadratic formula,
[tex]\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Using the above formula,
[tex]\text{ for } 3b^2 - 27b + 60 = 0 , \text{ we have } a = 3 ; b = -27 ; c = 60[/tex]
Substituting the values of a, b, c in above formula,
[tex]\begin{aligned}&b=\frac{-(-27) \pm \sqrt{(-27)^{2}-4(3)(60)}}{2 \times 3}\\\\&b=\frac{27 \pm \sqrt{729-720}}{6}\\\\&b=\frac{27 \pm \sqrt{9}}{6}\\\\&b=\frac{27 \pm 3}{6}\\\\&b=\frac{27+3}{6} \text { or } b=\frac{27-3}{6}\\\\&b=5 \text { or } b=4\end{aligned}[/tex]
Thus the required numbers are:
when b = 5,
a = 9 - b
a = 9 - 5
a = 4
Thus the number is 45
When b = 4,
a = 9 - b
a = 9 - 4
a = 5
Thus the number is 54
