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How many solutions are there to the equation x3 - 7 0?
None because without an equality sign the given expression is not an equation and so therefore no solutions are possible.
How many solutions are there in the quadratic equation x2 - 4x - 77 equals 0?
There are two solutions for x: x=11 and x=-7
What are the solutions to this equation 9x2 plus 6x-7 equals 0?
9x2+6x-7 = 0 Using the quadratic equation formula:- x = 0.6094757082 or x = -1.276142375
How do I work out 5xsquared plus 12x plus 7 equals 0?
5x2 + 12x + 7 = 0 This equation can be factored : (5x + 7) (x + 1) = 0 The equation equals zero when either 5x + 7 = 0 or x + 1 = 0 When 5x + 7 = 0 then 5x = -7 : x = -7/5 or -1.4 When x + 1 = 0 then x = -1 The solutions or roots of this equation are x = -7/5 and x = -1
How do you solve S squared plus 4s minus 21 equals to 0?
To solve the quadratic equation, S^2 + 4S - 21 = 0, you can factor the expression or use the quadratic formula. Factoring, we can rewrite it as (S-3)(S+7) = 0. This means that either S-3 = 0 or S+7 = 0. Solving for S in each case gives S = 3 or S = -7 as the solutions to the equation.
How many solutions are there to the equation x3 - 7 0?
None because without an equality sign the given expression is not an equation and so therefore no solutions are possible.
How many solutions are there in the quadratic equation x2 - 4x - 77 equals 0?
There are two solutions for x: x=11 and x=-7
Write a quadratic equation with these numbers -7 and -4 as solutions?
-3
What are the solutions to this equation 9x2 plus 6x-7 equals 0?
9x2+6x-7 = 0 Using the quadratic equation formula:- x = 0.6094757082 or x = -1.276142375
How do I work out 5xsquared plus 12x plus 7 equals 0?
5x2 + 12x + 7 = 0 This equation can be factored : (5x + 7) (x + 1) = 0 The equation equals zero when either 5x + 7 = 0 or x + 1 = 0 When 5x + 7 = 0 then 5x = -7 : x = -7/5 or -1.4 When x + 1 = 0 then x = -1 The solutions or roots of this equation are x = -7/5 and x = -1
How many solutions does the equation have?
The number of solutions an equation has depends on the nature of the equation. A linear equation typically has one solution, a quadratic equation can have two solutions, and a cubic equation can have three solutions. However, equations can also have no solution or an infinite number of solutions depending on the specific values and relationships within the equation. It is important to analyze the equation and its characteristics to determine the number of solutions accurately.
How do you solve S squared plus 4s minus 21 equals to 0?
To solve the quadratic equation, S^2 + 4S - 21 = 0, you can factor the expression or use the quadratic formula. Factoring, we can rewrite it as (S-3)(S+7) = 0. This means that either S-3 = 0 or S+7 = 0. Solving for S in each case gives S = 3 or S = -7 as the solutions to the equation.
How many solutions are there to the equation x2-7 equals 0?
There are 2 answers, +sqrt 7 and -sqrt7. Generally, the highest exponent in the expression (in this case 2) is the maximum number of possible answers.
Choose the quadratic equation with a leading coefficient of 1 that has as solutions -4 and 7?
x^2-3x-28=0...................
Write a quadratic equation in the variable x having the given numbers as solutions Type the equation in standard form ax2 plus bx plus c equals 0 Solution 7 is the on solution?
0x2 + 1x - 7 = 0
Why don't all rational equations have a single solution?
The number of solutions of a rational equation depends on the power (or degree) of the equation (that is, the highest power to which the variable is raised) and the domain. In the complex domain, each rational equation of power n has n solutions. It is, however, possible that two or more of these solutions are coincident - or "multiple zeros". In the real domain, the number of solutions can fall in pairs. So an equation of power 7 will always have 7 complex solutions but it can have 7, 5, 3 or 1 real solutions. (Real numbers are a subset of complex numbers). Another way of seeing this is through factorisation: The equation x3 + x2 - 10x + 8 = 0 can be factorised into (x + 1)*(x - 2)*(x - 4) = 0 Now the product of three numbers is 0 is any one of them is 0. That is, if x + 1 = 0 or if x - 2 = 0 or if x - 4 = 0 Thus the equation has the solutions: x = - 1, x = 2 or x = 4. The equation of order n can have at most n real binomial factors. (Any more and the biggest power of x would be bigger than n). And again, in the complex domain, using the binomial equation (and equivalents), it must have n binomial factors.
What is the product of the 2 solutions of the equation x'2 4'x-21 equals 0?
3 x - 7 = -21
