Expand: \( x^2 + x^2 + 6x + 9 = x^2 + 10x + 25 \)
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- \( S_{15} = \frac{15}{2} \times 52 = 15 \times 26 = 390 \)
- A right triangle has legs of lengths \( x \) and \( x+3 \), and a hypotenuse of length \( x+5 \). Find the value of \( x \).
- Using Pythagoras: \( x^2 + (x+3)^2 = (x+5)^2 \)
- \( 2x^2 + 6x + 9 = x^2 + 10x + 25 \)
- \( x^2 - 4x - 16 = 0 \)
- Use quadratic formula: \( x = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2} \)
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