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6x^2+4x=0 Divide everything by two 3x^2+2x=0 factor out an x x(3x+2)=0 Zero Product Property states that you can break apart the product when it is equal to zero because for this to occur either one or both of the factors must equal 0. x=0 or 3x+2=0 3x=-2 x=-2/3
x5
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
A quadratic equation has the form: x^2 - (sum of the roots)x + product of the roots = 0 or, x^2 - (r1 + r2)x + (r1)(r2) = 0
x10 = x5.x5 = (x5)2 [x power five whole squared] Equation is x10 + x5 - 2 Replacing x10 (x5)2+x5 - 2 Substituting x5 by Y Equation becomes Y2 + Y - 2 = 0 Y2 + 2Y - Y - 2 = 0 Y(Y + 2) - 1(Y+2) = 0 (Y-1)(Y+2) = 0 The values of Y are 1 and -2 Y = x5 = 1 Therefore, x = 1 Y = x5 = -2 x = fifth root of -2, which is an imaginary value.
x*-4=8+(2-5*x) =-4x=8+2-x5 x*-4=8+(2-5*x) =-4x=8+2-x5 x*-4=8+(2-5*x) =-4x=8+2-x5
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x5 - 2x4 - 24x3 = 0 x3(x2 - 2x - 24) = 0 x3(x2 + 4x - 6x - 24) = 0 x3(x - 6)(x + 4) = 0 x = {-4, 0, 6}
The product of the roots of the equation 2x2 -x -2 = 2 is 2x2 -x -2 = 2.
To round to estimate the product of 38 x 2, you first round each number to a more manageable value. In this case, you can round 38 to 40 and 2 to 0. Then, multiply the rounded numbers together to get an estimate of the product. In this case, 40 x 0 = 0. So, the estimated product of 38 x 2 is 0.
6x^2+4x=0 Divide everything by two 3x^2+2x=0 factor out an x x(3x+2)=0 Zero Product Property states that you can break apart the product when it is equal to zero because for this to occur either one or both of the factors must equal 0. x=0 or 3x+2=0 3x=-2 x=-2/3
In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.In theory, a quadratic equation can be separated into two factors. For example, in the equation x2 - 5x + 6 = 0, the left part can be factored as (x-3)(x-2) = 0. For the product to be zero, any of the two factors must be zero, so if either x - 3 = 0, or x - 2 = 0, the product is also zero. This gives you the two solutions.
x3/x1/2 = x5/2.
Product = Multiply7 x 0 = 0
3 x - 7 = -21
x5