The equation does not have real solutions, but it has pure imaginary numbers as solutions. -x^2 = 3 multiply by -1 to both sides x^2 = - 3 x = ± √-3 x = ± √[(-1)(3)] substitute i^2 for -1 x = ± √[(i^2)(3)] x = ± (√3)i
An equation with the solution set 1 and 3 can be written in factored form as (x-1)(x-3) = 0. When expanded, this equation becomes x^2 - 4x + 3 = 0. Therefore, the equation x^2 - 4x + 3 = 0 has the solution set 1 and 3.
5
Equation: x+3=3+x (notice the equal sign: equal; equation)Expression: x+3 (notice no equal sign)
Equation: x+3=3+x (notice the equal sign: equal; equation)Expression: x+3 (notice no equal sign)
It is: y = x+3
Move 3 over the right side of the equation so the equation would be x = -3. The graph of this would be a verticle line at x= -3
Y = - X + 3 You must zero out the Y - X + 3 = 0 - x = - 3 X = 3 ------- If you graphed the original equation you would see that.
A solution to an question makes the equation true. For example a solution to the equation 3x = x + 6 is x = 3, since 3(3) = 3+6.
(x - 3) and (x - (-5)).
This question cannot be answered without the whole equation: If the equation is y = 2x, then y = 6 for x = 3. If the equation is y = x + 2, then y = 5 for x = 3. If the equation is y = 99/x, then y = 33 for x = 3. If you can provide the complete equation, your question could be answered more precisely.
The equation is #of sides x (#of sides - 3) divided by 2. Equation: N x (N-3) 2