Question

One number is equal to the square of another. Find the numbers if both are positive and their sum is 380.

Answer 1

Answer:

19 and -20

Step-by-step explanation:

Call that number is x, we have:

$x + x^2 = 380\\(x^2 + 2*x*\frac{1}{2} + \frac{1}{4} ) -\frac{1521}{4} =0\\(x+\frac{1}{2})^2 =\frac{1521}{4} \\First situation: x + \frac{1}{2} = \sqrt{\frac{1521}{4} } =\frac{39}{2} \\x = \frac{39}{2} - \frac{1}{2} = 19\\Second situation: x + \frac{1}{2} =- \sqrt{\frac{1521}{4} } =-\frac{39}{2} \\\\x =- \frac{39}{2} - \frac{1}{2} = -20\\\\In conclusion, x= 19 and x=-20$Brainliest, please? It costs me so much time!