If one is the square of another number
and their sum is 12, calculate the
numbers

Question

If one is the square of another number and their sum is 12, calculate the numbers

UpStudy AI · Accepted Answer

Let the number be $$ x $$. According to the problem, one number is the square of another, so we can express this relationship as:

$$
x^2 = y
$$

where $$ y $$ is the other number. The problem also states that their sum is 12:

$$
x + y = 12
$$

Substituting $$ y $$ with $$ x^2 $$ in the sum equation gives us:

$$
x + x^2 = 12
$$

Rearranging this equation, we get:

$$
x^2 + x - 12 = 0
$$

Now, we can solve this quadratic equation using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 1 $$, $$ b = 1 $$, and $$ c = -12 $$. Plugging in these values:

$$
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}
$$
$$
x = \frac{-1 \pm \sqrt{1 + 48}}{2}
$$
$$
x = \frac{-1 \pm \sqrt{49}}{2}
$$
$$
x = \frac{-1 \pm 7}{2}
$$

This gives us two possible solutions for $$ x $$:

1. $$ x = \frac{6}{2} = 3 $$
2. $$ x = \frac{-8}{2} = -4 $$

Now, we can find the corresponding $$ y $$ values:

1. If $$ x = 3 $$:
   $$
   y = x^2 = 3^2 = 9
   $$
   So, the numbers are $$ 3 $$ and $$ 9 $$.

2. If $$ x = -4 $$:
   $$
   y = x^2 = (-4)^2 = 16
   $$
   So, the numbers are $$ -4 $$ and $$ 16 $$.

Thus, the pairs of numbers that satisfy the conditions are $$ (3, 9) $$ and $$ (-4, 16) $$.
The numbers are $$ 3 $$ and $$ 9 $$ or $$ -4 $$ and $$ 16 $$.

If one is the square of another number
and their sum is 12, calculate the
numbers

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