-x2 + 2x + 48 = (-x - 6)(x - 8)
3x2 + 6x - 24First factor out any common factors of the three terms. In this case, each term is a multiple of 3. So,3x2 + 6x - 24 = 3(x2 + 2x - 8)Next, use the reverse FOIL method to factor the polynomial in the parenthesis (x2 + 2x - 8). x2 factors out to x * x:(x + ??)(x + ??)Now, we must find two numbers that add to get 2 (from the 2x) and multiply to get -8. Those numbers are -2 and 4. So,(x2 + 2x - 8) = (x - 2)(x + 4)So,3x2 + 6x - 24 = 3(x - 2)(x + 4)
Possible example: 2|x| = x2 - 8 For x>=0, x2 - 2x - 8 = 0 (x-4)(x+2)=0 so x=4 (can't be -2 since x>=0) For x<0, x2 + 2x - 8 = 0 (x-2)(x+4)=0 so x=-4 (can't be 2 since x<0)
First, you'll need to find the x co-ordinates of the two points where that line and curve intersect:y = x2 + 2x - 3y = 4x + 45∴ x2 + 2x - 3 = 4x + 45∴ x2 - 2x - 48 = 0∴ (x - 8)(x + 6) = 0∴ x1 = -6, x2 = 8Now let's figure out which one of the two is higher on the y axis. This can be done by taking any point within the range and seeing which of the two equations has a higher value. Zero is between those points, and a fairly easy one to calculate, so let's try that:02 + 2·0 - 3 = -34·0 + 45 = 45So the line has the higher value. This means we'll need to take the area under the curve, and subtract it from the area under the line. This can be done by taking their definite integrals for the range -6 to 8, and subtracting the result for the curve from the result for the line.A = ∫-68 (4x + 25) dx - ∫-68 (x2 - 2x - 48) dx∴ A = (2x2 + 25x)|-68 - (x3/3 - x2 - 48x)|-68∴ A = [(2·64 + 25·8) - (2·36 + 25·-6)] - [(512/3 - 64 - 48·8) - (-216/3 - 36 - 48·-6)]∴ A = (128 + 200 - 72 - 150) - (170 2/3 - 64 - 384 +72 + 36 - 288)∴ A = 106 - (-457 1/3)∴ A = 563 1/3
7x+3-2x= if x=8 5x+3 5(8)+3 40+3 =43
I think this is how you do it: x2+2x = 8 x2+2x+(1/2*(2))2 = 8 + (1/2*(2))2 x2+2x+1=8+1 (x+1)2 = 9 SQ.ROOT [(x+1)2 = 9] x+1 = 3 and x+1 = -3 so, x=2 and x=-4
2x2 + 2x - 8 = 2(x2 + x - 4). This can take any value greater than or equal to -8.5
x2 + 2x - 48 = x2 + 8x - 6x - 48 = x(x + 8) - 6(x + 8) = (x - 6)(x + 8)
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
If you mean: x-3 = 2x-8 then the value of x works out as 5
-x2 + 2x + 48 = (-x - 6)(x - 8)
-x2 + 2x + 48 = (x +6)(8 - x)
Do you mean -x2-2x+8 = 0 Then if so the x intercepts are -4 and 2
-x2 + 2x + 48 = (-x - 6)(x - 8)
x3 + 8 = x3 + 23 = (x + 2)(x2 + 2x + 22) = (x + 2)(x2 + 2x + 4)
x + 7 / x - 2 = x + 4 / x - 3(times both sides by x - 3 and x - 2)(x + 7)(x - 3) = (x + 4)(x - 2)x2 + 7x - 3x - 21 = x2 + 4x - 2x - 8x2 + 4x - 21 = x2 + 2x - 8(- x2 from both sides)4x - 21 = 2x - 8(- 2x from both sides)2x - 21 = -8(+ 21 on both sides)2x = 13(divide both sides by 2)x = 6.5
X2 + 2X - 14 = 6X2 + 2X - 8 = 0what two products of - 8 = 2 ?(X +4)(X - 2)=========X = - 4X = 2