None because without an equality sign the given expression is not an equation and so therefore no solutions are possible.
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Since there are three numbers involved in the answer, the equation needs to have three solutions which makes it somewhat complicated. x3 + 87x2 + 1739x + 24360 = 0
2x2 - 6x - 25 = 0. Solutions are 5.34 and -2.34
A quadratic equation can have a maximum of 2 solutions. If the discriminant (b2-4ac) turns out to be less than 0, the equation will have no real roots. If the Discriminant is equal to 0, it will have equal roots. But, if the discriminant turns out to be more than 0,then the equation will have unequal and real roots.
A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)
As an equation, x/x = x2, or x2 = x/x. Multiplying both sides by x: x3 = x. Put everything on the left side: x3 - x = 0. Factoring the left part: x(x2 - 1) = 0, or x(x+1)(x-1) = 0. This has the solutions 0, 1, -1. However, 0 doesn't make sense in the original equation - it seems that multiplying with "x" introduced an extra solution.The remaining solutions, 1 and -1, fit the original specification.