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How do you solve the limit as x approaches zero of 1 over x minus 4 plus 1 over x plus 4 all over x?
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
How do you solve for the problem x squared minus 10 equals 4x plus 11?
x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
Which equation has the solution set 1 and 3?
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.
What is the sum of the roots x2 - 4x - 5 0?
Their sum is 4.
Can a quadratic equation have only one real solution?
Yes. x2-4x+4=0 (x-2)(x-2)=0 x=2
What is the solution of x2-4x-12 is less than 0?
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
What is the solution t0 x2-4x equals 0?
x2 - 4x = 0 x(x - 4) = 0 so either x = 0 or x - 4 = 0 that is, x = 0 or x = 4.
What is the solution to x2-4x-5 equals 0?
x = -1 or x = 5
What are solutions to the equation x2 plus 4x-9 equals 5x plus 3?
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
What is the answer to x2-4x-12 equals 0?
x = -2 and x = 6
X2 plus 4x - 9 equals 5x plus 3?
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
What is x2 plus 4x plus 4 equals 25?
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
What is the double solution for x2-4x plus a?
-2
X2 plus 4x-12 equals 0?
x2 + 4x - 12 = 0 so (x+6)*(x-2) = 0 so x = -6 or x = 2----To solve the quadratic equation x2 + 4x - 12 = 0, you can use two different methods:1: Reduce it to its factors:x2 + 4x - 12 = 0 is the same as (x + 6)(x - 2) = 0. But this can only be correct if one of the factors is zero which only happens when x = -6 or x = 2. That is the solution.2: Use the quadratic solution formula:x2 + 4x - 12 = 0 is of the form ax2 + bx + c = 0 with a = 1, b = 4, and c = -12.Substitute these values into the formulax = (-b ± sqrt(b2 - 4ac)) / 2a and you getx = (-4 ± sqrt(16 - (-48)) / 2= (-4 ±sqrt(64)) / 2= (-4 + 8) / 2 or (-4 - 8) / 2= 4 / 2 or -12 / 2so x = 2 or x = -6
Solve x2 - 4x - 9 equals 0 by completing the square?
x2 - 4x - 9 = 0 ∴ x2 - 4x = 9 ∴ x2 - 4x + 4 = 13 ∴ (x - 2)2 = 13 ∴ x - 2 = ±√13 ∴ x = 2 ± √13
What are the coordinates of the roots of x2 plus 4x equals 0?
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
X2 - 4x equals -4?
x2 - 4x + 4 = 0 (x - 2)(x - 2) = 0 x = 2
