If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
5x2 + 2x - 3 = 0 5x2 + 5x - 3x - 3 = 0 5x(x + 1) - 3(x + 1) = 0 (5x - 3)(x + 1) = 0 So 5x - 3 = 0 or x + 1 = 0 ie x = 3/5 or x = -1
5y2 + 5x2 = 1 Substitute y = 2x + 1 into this equation: 5*(2x + 1)2 + 5x2 = 1 5*(4x2 + 4x + 1) + 5x2 - 1 = 0 25x2 + 20x + 4 = 0 which factorises to (5x + 2)2 = 0 Therefore x = -2/5 And then y = 2x + 1 gives y = 1/5 So the two points of intersection are coincident, at (-2/5, 1/5)
It is 4x^2 + 10x - 1
-2x plus 3y equals 1
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
5x2 + 2x - 3 = 0 5x2 + 5x - 3x - 3 = 0 5x(x + 1) - 3(x + 1) = 0 (5x - 3)(x + 1) = 0 So 5x - 3 = 0 or x + 1 = 0 ie x = 3/5 or x = -1
5y2 + 5x2 = 1 Substitute y = 2x + 1 into this equation: 5*(2x + 1)2 + 5x2 = 1 5*(4x2 + 4x + 1) + 5x2 - 1 = 0 25x2 + 20x + 4 = 0 which factorises to (5x + 2)2 = 0 Therefore x = -2/5 And then y = 2x + 1 gives y = 1/5 So the two points of intersection are coincident, at (-2/5, 1/5)
x2 + 2x + 5 = 0x2 + 2x + 5 - 5 = 0 - 5x2 + 2x = -5x2 + 2x + (2/2)2 = -5 + (2/2)2x2 + 2x + 12 = -5 + 1(x + 1)2 = -4sq. root of (x + 1)2 = sq. root of -4|x + 1| = 2ix + 1 = +&- 2ix + 1 - 1 = -1 +&- 2ix = -1 +&- 2i
It is 4x^2 + 10x - 1
x=1
x = 0
-2x plus 3y equals 1
x=5.54x+1= 2x+12-1 -14x=2x+11-2x -2x2x=11/2 /2x=5.5
NM equals 2x + 1, as stated in the question!
Y = 2x + 1
The answer to 1 over 2x plus 13 equals 9 is x equals -8.