Limx→0 [ 1 / (x - 4) + 1 / (x + 4) ] / x = Limx→0 1 / (x2 - 4x) + 1 / (x2 + 4x) = Limx→0 (x2 + 4x) / (x4 - 16x2) + (x2 - 4x) / (x4 - 16x2) = Limx→0 (x2 + 4x - 4x + x2) / (x4 - 16x2) = Limx→0 2x2 / (x4 - 16x2) = Limx→0 2 / (x2 - 16) = 2 / (0 - 16) = -1/8
x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
This quadratic equation has no real roots because its discriminant is less than zero.
To determine the x-intercepts, i.e. the roots, of the equation y = x2 + 4x - 12, simply solve the quadratic equation x2 + 4x - 12 = 0.Recall that the roots of a quadratic equation Ax2 + Bx + C = 0 are given by the solution x = (-B +/- square-root (B2 - 4AC) / 2A.Plug the coefficients A=1, B=4, and C=-12 into this equation and you get.x = 4 and X = -12So the function crosses the X axis at x=4 and x=-12.Note: If the discriminant, B2 - 4AC, were zero, there would be one real root instead of two. If it were negative, there would be two complex conjugate roots.
To find which has imaginary roots, use the discriminant of the quadratic formula (b2 - 4ac) and see if it's less than 0. (The quadratic formula corresponds to general form of a quadratic equation, y = ax2 + bx + c)A) x2 - 1 = 0= 0 - 4(1)(-1) = 4Therefore, the roots are not imaginary.B) x2 - 2 = 0= 0 - 4(1)(-2) = 8Therefore, the roots are not imaginary.C) x2 + x + 1 = 0= 1 - 4(1)(1) = -3Therefore, the roots are imaginary.D) x2 - x - 1 = 0= 1 - 4(1)(-1) = 5Therefore, the roots are not imaginary.The equation x2 + x + 1 = 0 has imaginary roots.
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
x2-4x+4 = 0 (x-2)(x-2) = 0 x = 2 or x = 2 It has two equal roots.
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 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
Um, x2+3x-5=0? This is ax2+bx+c where a=1, b=3, and c=-5. The sum of the roots is -b/a so that means the sum of the roots is -3. Also, product of the roots is c/a. That means the product of the roots is -5. -3+(-5)= -8. There you have it.
x2 - 4x - 9 = 0 ∴ x2 - 4x = 9 ∴ x2 - 4x + 4 = 13 ∴ (x - 2)2 = 13 ∴ x - 2 = ±√13 ∴ x = 2 ± √13
First consider the auxiliary equation x2 - 4x - 12 = 0 This simplifies to (x + 2)(x - 6) = 0 which has root x = -2 and x = 6 Since the coefficient of x2 is positive, the inequality is satisfied between the roots that is, -2 < x < 6
x2 - 4x + 4 = 0 (x - 2)(x - 2) = 0 x = 2
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
Limx→0 [ 1 / (x - 4) + 1 / (x + 4) ] / x = Limx→0 1 / (x2 - 4x) + 1 / (x2 + 4x) = Limx→0 (x2 + 4x) / (x4 - 16x2) + (x2 - 4x) / (x4 - 16x2) = Limx→0 (x2 + 4x - 4x + x2) / (x4 - 16x2) = Limx→0 2x2 / (x4 - 16x2) = Limx→0 2 / (x2 - 16) = 2 / (0 - 16) = -1/8
x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4