x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
-2
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
x2 + 10x = 0 x2 + 10x + 25 = 25 (x + 5)2 = 25 x + 5 = +-5 x1 = 0 x2 =10
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)
no
-2
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3