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solve for x where x is a real number... x^2 - 2x - 15 < = 0
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\[x^2 - 2x - 15 \le 0\]
Solve for x over the real numbers: x^2-2 x-15 = 0 Factor the left hand side. The left hand side factors into a product with two terms: (x-5) (x+3) = 0 Solve each term in the product separately. Split into two equations: x-5 = 0 or x+3 = 0 Look at the first equation: Solve for x. Add 5 to both sides: x = 5 or x+3 = 0 Look at the second equation: Solve for x. Subtract 3 from both sides: Answer: | | x = 5 or x = -3
@zkrup what if it's \[x^4 -1 / x^3 = 0\]
how would I do that? @zkrup
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