If you mean: x2+8x-9 = 0 then the solutions are x = 1 and x = -9
They are x = -1 and x = 4.
Solutions: x = 9 and x = 1 Factored: (x-9(x-1) = 0 Equation: x2-10x+9 = 0
If the equation is x2 = k0 Then there are two solutions: + or - 1, since any number to 0 is 1 and 1 or -1 squared = 1.
Two cases in which this can typically happen (there are others as well) are: 1. The equation includes a square. Example: x2 = 25; the solutions are 5 and -5. 2. The equation includes an absolute value. Example: |x| = 10; the solutions are 10 and -10.
If you mean: x2+8x-9 = 0 then the solutions are x = 1 and x = -9
They are x = -1 and x = 4.
Solutions: x = 9 and x = 1 Factored: (x-9(x-1) = 0 Equation: x2-10x+9 = 0
If the equation is x2 = k0 Then there are two solutions: + or - 1, since any number to 0 is 1 and 1 or -1 squared = 1.
Oh, dude, for this quadratic equation x^2 - 5x + 4 = 0, the solutions are just the roots of the equation. You can find them by either factoring the quadratic or using the quadratic formula. So, like, the solutions are x = 1 and x = 4. Easy peasy lemon squeezy!
It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.
Two cases in which this can typically happen (there are others as well) are: 1. The equation includes a square. Example: x2 = 25; the solutions are 5 and -5. 2. The equation includes an absolute value. Example: |x| = 10; the solutions are 10 and -10.
This is a quadratic equation which will have two solutions: X2 = 4x+5 Rearrange the equation: x2-4x-5 = 0 Factor the equation: (x+1)(x-5) = 0 So the solutions are: x = -1 or x = 5
x = 2 and x = -5
Nope. Consider x2+0x+1=0. This means x2+1=0. This has two solutions, but they are complex numbers: +i and -i, where i is the squareroot of -1. How about x2+0x+0=0? This means x2=0. This has two solutions, sure, but they aren't distinct. In this case, x=0 for both solutions, so we just consider them one solution.
x2 -y2 =16 This is an equation that describes your problem. We can write this equation as (1/16)x2 -(1/16)y2 =1 You may recognize this as the equation whose graph is a hyperbola. So there are an infinite number of solutions.
equation1: y+1 = x => y = x-1 equation 2: y = x2-6x++5 If, x2-6x+5 = x-1 then, x2-7x+6 = 0, solving this quadratic equation, x = 1 or x = 6 Substituting into equation 1: when x = 1, y = 0 and when x = 6, y = 5 Therefore the solutions are: x = 1, y = 0 and x = 6, y = 5