To solve the equation (x^2 + y^2 = 100), you can express it in terms of (y) as (y = \pm\sqrt{100 - x^2}). This represents a circle with a radius of 10 centered at the origin in the Cartesian plane. You can find specific solutions by substituting different values for (x) within the range of -10 to 10, and calculating the corresponding (y) values.
(2x)ysquared
(2-r)e-rr
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
(4 ± i2) where i2 = -1
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
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)
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
(2x)ysquared
(2-r)e-rr
X2 + y2 = 100 = r2 r = 10
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
x=4 x=0
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
(4 ± i2) where i2 = -1
x2 + x2 = 2x2
x2+7x+12
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}