x^2 - 3x - 1 = 0 x = -b+-root(b^2 - 4ac)/2a x = 3 +-root(9+4)/2 Two solutions: 1. x = (3+root(13))/2 2. x = (3-root(13))/2
The square root of 1 is 1.The square root of 0 is 0.
This is a quadratic equation in the form of x2-2x-1 = 0 which will have two solutions and can be solved using the quadratic equation formula: x = -b/2a + and - the square root of (b2-4ac)/2a Whereas in this equation a = 1, b = -2 and c = -1 Solutions: x = 1-the square root of 2 or x = 1+the square root of 2 Remember that (- -2) = 2, (-2)2 = 4 and (-4*1*-1) = 4
You need to be more specific. A quadratic equation will have 2 solutions. The 2 solutions can be equal (such as x² + 2x + 1 = 0, solution is -1 and -1). If one of the solutions is a real number, then the other solution will also be a real number. If one of the solutions is a complex number, then the other solution will also be a complex number. [a complex number has a real component and an imaginary component]In the equation: Ax² + Bx + C = 0. The term [B² - 4AC] will determine if the solution is a double-root, or if the answer is real or complex.if B² = 4AC, then a double-root (real).if B² > 4AC, then 2 real rootsif B² < 4AC, then the quadratic formula will produce a square root of a negative number, and the solution will be 2 complex numbers.If B = 0, then the numbers will be either pure imaginary or real, and negatives of each other [ example 2i and -2i are solutions to x² + 4 = 0]Example of 2 real and opposite sign: x² - 4 = 0; 2 and -2 are solutions.
x2 + 3x + 3 = 0 doesn't factor neatly. Applying the quadratic formula we find two imaginary solutions: (-3 plus or minus the square root of -3) divided by 2x = -1.5 + 0.8660254037844386ix = -1.5 - 0.8660254037844386iwhere i is the square root of negative 1
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.
x^2 - 3x - 1 = 0 x = -b+-root(b^2 - 4ac)/2a x = 3 +-root(9+4)/2 Two solutions: 1. x = (3+root(13))/2 2. x = (3-root(13))/2
4x2 - 4x + 1 = 0 => (2x - 1)(2x - 1) = 0 => (2x - 1)2 = 0 There is one solution: x = 1/2. It is a repeated root of the equation.
The two solutions are the conjugate complex numbers, +i*sqrt(11) and -i*sqrt(11), where i is the imaginary square root of -1.
The square root of 1 is 1.The square root of 0 is 0.
This is a quadratic equation in the form of x2-2x-1 = 0 which will have two solutions and can be solved using the quadratic equation formula: x = -b/2a + and - the square root of (b2-4ac)/2a Whereas in this equation a = 1, b = -2 and c = -1 Solutions: x = 1-the square root of 2 or x = 1+the square root of 2 Remember that (- -2) = 2, (-2)2 = 4 and (-4*1*-1) = 4
You need to be more specific. A quadratic equation will have 2 solutions. The 2 solutions can be equal (such as x² + 2x + 1 = 0, solution is -1 and -1). If one of the solutions is a real number, then the other solution will also be a real number. If one of the solutions is a complex number, then the other solution will also be a complex number. [a complex number has a real component and an imaginary component]In the equation: Ax² + Bx + C = 0. The term [B² - 4AC] will determine if the solution is a double-root, or if the answer is real or complex.if B² = 4AC, then a double-root (real).if B² > 4AC, then 2 real rootsif B² < 4AC, then the quadratic formula will produce a square root of a negative number, and the solution will be 2 complex numbers.If B = 0, then the numbers will be either pure imaginary or real, and negatives of each other [ example 2i and -2i are solutions to x² + 4 = 0]Example of 2 real and opposite sign: x² - 4 = 0; 2 and -2 are solutions.
0
x2 + 3x + 3 = 0 doesn't factor neatly. Applying the quadratic formula we find two imaginary solutions: (-3 plus or minus the square root of -3) divided by 2x = -1.5 + 0.8660254037844386ix = -1.5 - 0.8660254037844386iwhere i is the square root of negative 1
The quadratic equation provides solutions for the generalized equation ax2 + bx + c = 0. The solution(s) are (-b +/- square root ( b2 - 4ac)) / 2a. Plugging in a=1, b=3, and c=-15 (from x2 + 3x -15 = 0) you get... (-3 +/- square root (32 - 4(1)(-15)) / 2(1) ... or ... (-3 +/- square root (9 + 60)) / 2 ... or ... (-3 +/- square root (69)) / 2 ... or ... about -3 +/- 8.3 Since 69 is positive, the square root (69) is real, at 8.3 This equation has two real roots, x=5.3 and x=-11.3 If the discriminant (b2 - 4ac) were zero, there would be one real root. If it were negative, there would be one real root and one imaginary root, i.e. a complex conjugate.
There is no equals sign visible but, taking the equation to be8x2 + 2x + 1 = 0,the solutions are the complex conjugate pair, -1.25 ± 0.330719*i where i is the imaginary square root of -1.
x2+10x+1 = -12+2x x2+8x+13 = 0 The solution: x = -4 + the square root of 3 or x = -4 - the square root of 3