I suggest you apply the quadratic formula (a = 2, b = -4, c = -1).
x2-4x+4 = 0 (x-2)(x-2) = 0 x = 2 or x = 2 It has two equal roots.
x2+4x+4 = 0 (x+2)(x+2) = 0 x = -2 and also x = -2 When the solutions to a quadratic equation are identical this means that both values are at the turning point or vertex of the parabola just on the x axis. d>=0,then roots of quadratic equation are real if D<0,then roots are imaginary
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
To solve the equation (4x + 8 - 20 = 0), first simplify it: (4x - 12 = 0). Then, add 12 to both sides to get (4x = 12). Finally, divide by 4 to find (x = 3).
To find the roots of the equation (-x^2 - 4x + 12 = 0), we can first rewrite it as (x^2 + 4x - 12 = 0). Using the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) with (a = 1), (b = 4), and (c = -12), we calculate the discriminant: (b^2 - 4ac = 16 + 48 = 64). Thus, the roots are (x = \frac{-4 \pm 8}{2}), giving us (x = 2) and (x = -6).
2x2 -3x - 4 = 0 use the qudratic equation and your answers are 1.137458609 and -2.637458609
The product of the roots of the equation 2x2 -x -2 = 2 is 2x2 -x -2 = 2.
There are 2 roots to the equation x2-4x-32 equals 0; factored it is (x-8)(x+4); therefore the roots are 8 & -4.
x2-4x+4 = 0 (x-2)(x-2) = 0 x = 2 or x = 2 It has two equal roots.
Their sum is 4.
x2+4x+4 = 0 (x+2)(x+2) = 0 x = -2 and also x = -2 When the solutions to a quadratic equation are identical this means that both values are at the turning point or vertex of the parabola just on the x axis. d>=0,then roots of quadratic equation are real if D<0,then roots are imaginary
The equation must have roots of x = -1 and x = 5 So: x + 1 = 0 and x - 5 = 0 Therefore: (x + 1)(x - 5) = 0 Expanding the brackets gives the equation: x2 - 4x - 5 = 0
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
To solve the equation (4x + 8 - 20 = 0), first simplify it: (4x - 12 = 0). Then, add 12 to both sides to get (4x = 12). Finally, divide by 4 to find (x = 3).
x2 - 4x - 5 = zero( x - 5 ) ( x + 1 ) = zerox - 5 = zero ...... X = 5x + 1 = Zero ...... X = -1
To find the roots of the equation (-x^2 - 4x + 12 = 0), we can first rewrite it as (x^2 + 4x - 12 = 0). Using the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) with (a = 1), (b = 4), and (c = -12), we calculate the discriminant: (b^2 - 4ac = 16 + 48 = 64). Thus, the roots are (x = \frac{-4 \pm 8}{2}), giving us (x = 2) and (x = -6).
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.