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Module Progress: Modulus of Trigonometric Functions

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Modulus of Trigonometric Functions

question 1 of 2 Grade: Ten - G10

Q: Simplify | -7 | + | 3 |.

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Half of 8 is 4, square is 16. Add and subtract 16: \(x^2 + 8x + 16 - 16 + 5 = 0\), so \((x+4)^2 - 11 = 0\), \((x+4)^2 = 11\), \(x+4 = \pm\sqrt{11}\), \(x = -4 \pm \sqrt{11}\).

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Did You Know?

Half of 8 is 4, square is 16. Add and subtract 16: \(x^2 + 8x + 16 - 16 + 5 = 0\), so \((x+4)^2 - 11 = 0\), \((x+4)^2 = 11\), \(x+4 = \pm\sqrt{11}\), \(x = -4 \pm \sqrt{11}\).

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