It is known that the two roots of the quadratic equation (6-k) (9-k) X2 - (117-15k) x + 54 = 0 with respect to X are integers, and the values of all real numbers K satisfying the conditions are obtained

It is known that the two roots of the quadratic equation (6-k) (9-k) X2 - (117-15k) x + 54 = 0 with respect to X are integers, and the values of all real numbers K satisfying the conditions are obtained

(6-k) X-9] [(9-k) X-9 (9-k) X-6] [(9-k) x-9-9 (6-k-k) X-9] [(9-k-k) X-9 (9-k-k) X-9 (6-k) X-9 (6-k) x-9-9-9-9-k-9-9-k-9-k-9-k-9-k-6] = 0. Because this equation is a quadratic equation about X, so, K ≠ 6, K ≠ 6, k = 6X1 (6-k) is a quadratic = 6X1 = 6x1-6x1-6x1-6x1-6x1-9x1-9x1-9x1 = 9x1 = 9x2-6x2 = 9x2 = 9x2-6x2 = 9x2-6x2-6x2 = 9x2-6x2 = 9x2 = 9x2 = 9x2 x2 x2 = 9x2 x2 , − 1, 1, 2, 3, 6x2 − 2 = 1, 2, 3, 6, − 6, − 3, − 2, − 1. So X1 = - 9, - 6, - 5, - 4, - 2, - 1, 0, 3. And K = 6X1 − 9x1 = 6-9x1. Substituting X1 = - 9, - 6, - 5, - 4, - 2, - 1, 3 respectively, we get k = 7152395334212, 15, 3

It is known that the two roots of the quadratic equation (6-k) (9-k) X2 - (117-15k) x + 54 = 0 with respect to X are integers, and the values of all real numbers K satisfying the conditions are obtained

The original equation can be reduced to: [(6-k) X-9] [(9-k) X-6] = 0. Because this equation is a quadratic equation of one variable with respect to x, K ≠ 6, K ≠ 9, then there are: X1 = 96 − K (1), X2 = 69 − K (2)

B ^ 2-4ac = k ^ 2-8, when k is the value, the equation has two unequal real roots
, 2. There are two equal real roots, 3. There are real roots, 4. There are no real roots

∵⊿＝b²－4ac＝k²－8
(1) There are two unequal roots of real numbers ﹥ 0 ﹥ K & ᦇ 178; ﹥ 8 ﹥ K ﹥ 2 √ 2 or K ﹤ 2 √ 2
(2) There are two equal roots of real numbers ⊿ = 0 ⊿ K & # 178; = 8 ⊿ k = ± 2 √ 2
(3) There are roots of real numbers ≠ ⊿ ≥ 0 ﹥ K & ᦇ 178; ≥ 8 ﹥ K ≥ 2 √ 2 or K ≤ - 2 √ 2
(4) There is no real number root ⊿ 0 ⊿ K & # 178; < 8 ⊿ - 2 √ 2 < K < 2 √ 2

Write the root of the following equation directly: 4x-1 = 0:_____ Write the root of the following equation directly: the square of X - 2x - 3 = 0______

Write the root of the following equation directly: 4x-1 = 0:__ x=1/4___ Write the root of the following equation directly: the square of X - 2x - 3 = 0__ x=3,x=-1____

Given two points a (0, 1), B (2, m), if there is only one circle passing through a and B and tangent to the X axis, we can find the value of M and the equation of the circle

Let the round equation be (x-a) 2 + (y-b) 2 = B2, then there is A2 + (1-B) 2 = B2 (2-A) 2 (2-A) 2 (2-A) 2 + (M-B) 2 = B2, we can get: (1-m) a2-4a + 4 + M2-4 + m2-m2-m = 0, (3 points) when m ≠1, (3 points) m (3) m ≠ 1, we can get: m (m2-2-2m-2m-2-2m + 5) = 0, so when m = 0 = 0 = 0 = 0 = 0 m = 0 m = 1, a = 1, so B = 1, the equation is (x-1) 2 + (Y-1) 2 (x-1) 2 + (Y-1) 2 (Y-1) 2 + (y-1-1) 2 (y-1-1) 2 (Y-1) 2 + (y-1-1) (y-1-1-1-1-1) 2 = 1, M = 0 The equation of circle is (X-2) 2 + (y-52) 2 = 254

1 passes through points a (0,1) and B (4, m), and there is only one point tangent to the X axis. Find the value of M and the equation of the corresponding circle

Let the equation of O (a, b) circle be: (x-a) square + (y-b) square = R square
Because it is tangent to the x-axis, then r = 1, that is: (x-a) square + (y-b) square = 1, substituting points (0,1), (4, m) into the square,
There are: (0-A) square + (1-B) square = 1
(4-A) square + (M-B) square = 1=

Given that the image of the function y = x2 + (M + 4) x-2m-12 intersects with the X axis at two points and is on the right side of the point (1,0), then the value range of M is______ ．

If a and B are on the right side of the straight line x = 1, let a (a, 0), B (B, 0), then a > 1, b > 1, then (a − 1) + (B − 1) > 0 (a − 1) (B − 1) > 0, the solution is: M > 4, known from △ 0, M > 20, so the answer is: M > 20

Y = x square + (M + 4) x-2m-12 and X axis intersect at two points, which are on the right side of point (1,0). Find the value range of real number M

y=x^2+（m+4）x-2m-12
=[x+(m+6)](x-2)
Let y = 0,
The solution is X1 = 2, X2 = - M-6
∵ function and X axis intersect at two points, which are on the right side of point (1,0)
∴-m-6>1
∴m

Let the distance difference between point P and point (- 1,0) and (1,0) be 2m, and the distance ratio between point P and X axis and Y axis be 2

The coordinate of point P (x, y) is (x, y) and let 124124124124124124124124124124x = 2, that is, y = ± 2x, X ≠ 0, therefore, three points P (x, y, y, y, m (x, y), m (- 1,0), m (- 1,0), n (1,0 sox2m2 − y21 − M2 = 1. Substitute y = ± 2x into x2m2 − y21 − M2 = 1 It is obtained that x2 = M2 (1 − m2) 1 − 5m2 ≥ 0, because 1-m2 > 0, so 1-5m2 > 0, the solution is 0 | m | 55, that is, the value range of M is (− 55, 0) ∪ (0, 55)

Given that there is only one circle passing through points a (0, 1), B (4, a) and tangent to X axis, the value of a and the equation of the corresponding circle are obtained

Consider two cases: (I) let the center coordinate of the circle be (x, y), when point B is the tangent point, B is on the x-axis, so a = 0. Then B (4, 0), so the midpoint coordinate of AB is (2, 12), and the slope of the straight line AB is 1 − 00 − 4 = - 14, then the slope of the vertical line AB is 4, so the equation of the vertical line AB is Y-12 = 4 (X-2) and x = 4, so the equation of the circle is: (x-4) 2 + (Y − 172) 2 = (172) 2; (II) when a = 1, AB is parallel to X axis, then the vertical equation of AB is x = 2, let the center coordinate of the circle be (2, y), according to Pythagorean theorem: y2 = 22 + (Y-1) 2, the solution is y = 52, so the equation of the circle is: (X-2) 2 + (Y − 52) 2 = (52) 2 The range is: (X-2) 2 + (Y − 52) 2 = (52) 2