Let the two consecutive positive integers be $x$ and $x+1$. Their squares add up to 630:

$x^2 + (x+1)^2 = 630$

Step 1: Expand the equation

$x^2 + (x^2 + 2x + 1) = 630$

$2x^2 + 2x + 1 = 630$

Step 2: Simplify the equation

$2x^2 + 2x - 629 = 0$

Step 3: Divide by 2 to simplify further

$x^2 + x - 314.5 = 0$

Step 4: Solve using the quadratic formula

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 1$, and $c = -314.5$. Substituting these values:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-314.5)}}{2(1)}$

$x = \frac{-1 \pm \sqrt{1 + 1258}}{2}$

$x = \frac{-1 \pm \sqrt{1259}}{2}$

Step 5: Approximate $\sqrt{1259}$

$\sqrt{1259} \approx 35.48$

$x = \frac{-1 \pm 35.48}{2}$

Step 6: Calculate the two solutions

$x = \frac{-1 + 35.48}{2} = \frac{34.48}{2} \approx 17.24$

$x = \frac{-1 - 35.48}{2} = \frac{-36.48}{2} \approx -18.24 \, \text{(not positive)}$

Since $x$ must be an integer, the closest value is $x = 17$.

Step 7: Verify the solution

The integers are $17$ and $18$. Their squares are:

$17^2 + 18^2 = 289 + 324 = 613$

Therefore, the two consecutive positive integers are 17 and 18.

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Related Questions

1. How do you derive the quadratic formula?
2. What are other methods to solve quadratic equations?
3. How can we check if a number is a perfect square?
4. How to solve a system of equations involving consecutive integers?
5. How can we extend this to sums of cubes?

Tip: Always verify your solution by substituting it back