MHB How to solve this dreadful quadratic with paper and pen?

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To solve the equation 600*sqrt{1-a^2/400}=578-a^2/2, one effective method is to substitute a^2 with x and square both sides, resulting in quadratic polynomials. Simplifying the left-hand side to 30*sqrt{400-a^2} can streamline the process. Multiplying both sides by 2 helps eliminate fractions, making calculations easier. It's crucial to check that x remains less than or equal to 400 to avoid extraneous roots from squaring. Overall, these strategies can lead to a more elegant solution.
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600*sqrt{1-a^2/400}=578-a^2/2

Shorter and elegant tricks would be welcome!
 
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The straigtforward way is to denote $a^2$ by $x$ and take the square of both sides. Then both sides become quadratic polynomials. After finding $x$, it is necessary to check that $x\le 400$ to avoid gaining extra roots during squaring. The left-hand side can be simplified to $30\sqrt{400-a^2}$; then both sides can be multiplied by 2 to avoid fractions. It may help a little to denote $400-x$ by $y$.
 
Evgeny.Makarov said:
The straigtforward way is to denote $a^2$ by $x$ and take the square of both sides. Then both sides become quadratic polynomials. After finding $x$, it is necessary to check that $x\le 400$ to avoid gaining extra roots during squaring. The left-hand side can be simplified to $30\sqrt{400-a^2}$; then both sides can be multiplied by 2 to avoid fractions. It may help a little to denote $400-x$ by $y$.

Thanks :)
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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