(2x)ysquared
(2-r)e-rr
(4 ± i2) where i2 = -1
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -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)
(2x)ysquared
(2-r)e-rr
X2 + y2 = 100 = r2 r = 10
x=4 x=0
x2 + 20x +0 =30 [(20/2)2 =100] x2 + 20x + 100 =30 +100 √(x+10)2=√130 x+10=√130 x= −10+√130 √ means square root
(4 ± i2) where i2 = -1
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + x2 = 2x2
x2+7x+12
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}