Question

One number is the square of another. their sum is 182. find the numbers.

Answer 1

$n-first\ number\\n^2-second\ number\\\\\text{The equation:}\\\\n+n^2=182\qquad\text{subtract 182 from both sides}\\\\n^2+n-182=0\\\\n^2+14n-13n-182=0\\\\n(n+14)-13(n+14)=0\\\\(n+14)(n-13)=0\iff n+14=0\ \vee\ n-13=0\\\\\boxed{n=-14\ or\ n=13}\\\\n^2=(-14)^2=196\ ot\ n^2=13^2=169\\\\Answer:\ \boxed{196\ and\ -14\ or\ 169\ and\ 13}$

Answer 2

Taking into account the definition of a system of linear equations, the numbers can be:

• 13 and 169
• -14 and 196

System of linear equations

A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.

Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values ​​of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.

This case

In this case, a system of linear equations must be proposed taking into account that "x" and "y" are two numbers.

One number is the square of another, and the sum of the two numbers is 182. So, the system of equations to be solved is

$\left \{ {{x=y^{2} } \atop {x+y=182}} \right.$

There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.

In this case, substituting the first equation in the second you get:

y² + y= 182

Solving:

y² + y- 182= 0

Being this a quadratic function of the form ax² + by +c, then it can be solved by: $x1,x2=\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

In this case, being a=1, b=1 and c=-182, the equation is solved by:

$y1=\frac{-1+\sqrt{1^{2} -4x1x(-182)} }{2x1}$

$y1=\frac{-1+\sqrt{1 +728} }{2}$

$y1=\frac{-1+\sqrt{729} }{2}$

$y1=\frac{-1+27 }{2}$

$y1=\frac{26 }{2}$

y1= 13

Remembering that x=y², then x1=13². So x1= 169.

and

$y2=\frac{-1-\sqrt{1^{2} -4x1x(-182)} }{2x1}$

$y2=\frac{-1-\sqrt{1 +728} }{2}$

$y2=\frac{-1-\sqrt{729} }{2}$

$y2=\frac{-1-27 }{2}$

$y2=\frac{-28 }{2}$

y2= -14

Remembering that x=y², then x2=(-14)². So x1= 196.

Finally, the numbers can be:

• 13 and 169
• -14 and 196

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