Answer:
the number can be either 8 or 3.
Step-by-step explanation:
Let's define N as the "number"
We know that when the square of this number is increased by 24:
N^2 + 24
we got eleven times the original number, then:
N^2 + 24 = 11*N
We just need to solve this for N
To do it, we first move all the terms to one side of the equation:
N^2 - 11*N + 24 = 0
Now we can use the Bhaskara's formula for the zeros of a quadratic equation:
for a general quadratic equation:
a*x^2 + b*x + c = 0
the roots or zeros are given by:
[tex]x = \frac{-b \pm \sqrt{b^2 - 4*a*c} }{2*a}[/tex]
We get then:
[tex]N = \frac{-(-11) \pm \sqrt{(-11)^2 - 4*1*24} }{2*1} = \frac{11 \pm 5}{2}[/tex]
So we have two solutions:
N = (11 + 5)/2 = 16/2 = 8
N = (11 - 5)/2 = 6/2 = 3
So the number can be either 8 or 3.
