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
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.
2x2 - 6x - 25 = 0. Solutions are 5.34 and -2.34
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.
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)
0 = 0 is an identity and not an equation. Equations have solutions, identities do not.
If the discriminant of a quadratic equation is less then 0 then it will have no real solutions.
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
It has infinitely many solutions.
Factor the equation and solve for x in each part. x3-4x2-12x Factor out an x right away x (x² - 4x - 12) factor the trinomial x (x - 6) (x + 2) now solve each for y=0 x=0, (x-6=0), (x+2=0) The zeroes (solutions) to the equation are {-2 0 6 }■
There are infinitely many solutions to the equation since it simplifies to 13 = 13, which is always true.
If you mean: 3x^2+12x+12 = 0 then it has two equal solutions of -2
There are no real solutions because the discriminant of the quadratic equation is less than zero.
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.
2x2 - 6x - 25 = 0. Solutions are 5.34 and -2.34
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.
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)