Solve the following equations 1, x2-4x-9 = 0 2, X2 + 6x-2 = 0
(1)x²-4x-9=0
x²-4x+4-13=0
(x-2)²=13
x1=2-√13,x2=2+√13
(2)x²+6x-2=0
x²+6x+9-11=0
(x+3)²=11
x1=-3-√11,x2=-3+√11
When √ 2-x is meaningful, simplify the result of √ x2-4x + 4 - √ x2-6x + 9
So √ (X & # 178; - 4x + 4) = √ (X-2) 178; = | X-2 | = 2-x and √ (X & # 178; - 6x + 9) = √ (x-3) 178; = | x-3 | = 3-x, so √ (X & # 178; - 4X + 4) - √ (X & # 178; - 6x + 9) = 2-x - (3-x) = - 1
Common tangent of two circles x2 + y2 = 1 and X2 + y2-6x-8y + 9 = 0
How many common tangent lines are there
X & # 178; + Y & # 178; = 1, the center of the circle is (0,0), the radius r = 1 x & # 178; + Y & # 178; - 6x-8y + 9 = 0, that is, (x-3) 178; + (y-4) 178; = 16, the center of the circle is (3,4), the radius r = 4, the center distance of the circle is √ [(3-0) 178; + (4-0) 178;] = 5 = R + R, there are three common tangents (one inner common tangent and one outer common tangent)