y=2x2-3x2-12x+5=0
3x2 + 12x = - 1 ie 3x2 + 12x + 1 = 0 has no rational roots. The irrational roots are [-12 +/- sqrt(132)]/6 = -3.915 and -0.085
-2
1 and 3
Given, 3x2 - 4x = -2 Then, 9x2 - 12x = -6; 9x2 - 12x + 4 = -2 {completing the square} ; 3x - 2 = ±i√2 {sq rt of both sides} ; and 3x = 2 ± i√2. Therefore, x = ⅓(2 ± i√2).
y=2x2-3x2-12x+5=0
It is x = +/- 2 depending on whether the second term in the equation is -12x or +12x.
3x2 + 12x = - 1 ie 3x2 + 12x + 1 = 0 has no rational roots. The irrational roots are [-12 +/- sqrt(132)]/6 = -3.915 and -0.085
GCF = 12x.24x = 2(12x)36x3 = 3x2(12x)
3x2 -12x+24=-10x-20-3x2+6
The equation is equivalent to: -3x2 - 12x - 1 = 0 [you could change signs to get 3x2 + 12x + 1 = 0 but that is not required] The discriminant is (-12)2 - 4*(-3)(-1) = 144 - 12 = 132
-2
1 and 3
If you mean: 3x^2+12x+12 = 0 then it has two equal solutions of -2
an upside down parabola
Given, 3x2 - 4x = -2 Then, 9x2 - 12x = -6; 9x2 - 12x + 4 = -2 {completing the square} ; 3x - 2 = ±i√2 {sq rt of both sides} ; and 3x = 2 ± i√2. Therefore, x = ⅓(2 ± i√2).
Every polynomial defines a function, often called P. Linear and and quadratic function belong to a family of functions known as polynomial functions, which often are called P(x). When P(x) = 0, we call it an equation. Any value of x for which P(x) = 0 is a root of the equation and a zero of the function. Polynomials of the first few degrees have a special names such as:Degree 0: Constant functionDegree 1: Linear functionDegree 2: Quadratic functionDegree 3: Cubic functionDegree 4: Quartic functionDegree 5: Quintic functionSo, if we work a little bit to the given expression, we can turn it in a polynomial function of the second degree.y - 3x^2 = 12x - 7y - 3x^2 + 3x^2 = 12x - 7 + 3x^2y = 3x^2 + 12x - 7Let's write y = f(x) and f(x) = 3x^2 + 12x - 7, where a = 3, b = 12, and c = -7.Since a > 0, the parabola opens upward, so we have a minimum value of the function. The maximum or minimum value of the quadratic function occurs at x = -(b/2a).x = -12/6 = -2To find the minimum value of the function, which is also the y-value, we will find f(-2).f(-2) = 3(-2)^2 + 12(-2) - 7f(-2) = 12 - 24 - 7 = -19Thus the minimum value of the function is -19, and the vertex is (-2, -19)To find zeros, we solve f(x) = 0. So,f(x) = 3x^2 + 12x - 7f(x) = 03x^2 + 12x - 7 = 0 In order to solve this equation by completing the square, we need the constant term on the right hand side;3x^2 + 12x = 7 Add the square of one half of the coefficient of x to both sides, (6^2)3x^2 +12x + 36 = 7 + 36 Use the formula (a + b)^2 = a^2 + 2ab + b^2;(3x + 6)^2 = 43 Take the square root of both sides, and solve for x;3x + 6 = (+ & -)square root of 433x + 6 = (+ & -)(square root of 43) Subtract 6 to both sides;3x = (+ & -)(square root of 43) - 6 Divide both sides by 3;x = (square root of 43)/3 - 2 or x = -(square root of 43)/3 - 2The solution are (square root of 43) - 2 and -(square root of 43) - 2