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∙ 9y agoNone because without an equality sign the given expression is not an equation and so therefore no solutions are possible.
There are two solutions for x: x=11 and x=-7
9x2+6x-7 = 0 Using the quadratic equation formula:- x = 0.6094757082 or x = -1.276142375
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
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.
None because without an equality sign the given expression is not an equation and so therefore no solutions are possible.
There are two solutions for x: x=11 and x=-7
-3
9x2+6x-7 = 0 Using the quadratic equation formula:- x = 0.6094757082 or x = -1.276142375
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? 4x+1=5+2(2-4) a. one solution b. infinite solutions c. no solution
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.
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.
x^2-3x-28=0...................
0x2 + 1x - 7 = 0
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.
3 x - 7 = -21