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It is known that m and N are the two real roots of the equation x-4x + 3 = 0, and m
Because x-4x + 3 = 0, (x-1) (x-3) = 0 because M
It is known that as shown in Figure 13m and N are two real roots of equation x2-6x + 5 = 0, and m < n, the image of parabola y = - x2 + BX + C passes through points a (m, 0) and B (0, n) ① Let the other intersection point of the parabola and the x-axis be C and the vertex of the parabola be d. try to find the coordinates of points c and D and the area of △ BCD; (Note: the vertex coordinates of the parabola y = AX2 + BX + C (a ≠ 0) are (− B2A, 4ac − b24a))
(1) Solve the equation x2-6x + 5 = 0, get X1 = 5, X2 = 1, from m < n, there is m = 1, n = 5, so the coordinates of point a and B are a (1, 0), B (0, 5) respectively. Substitute the coordinates of a (1, 0), B (0, 5) into y = - x2 + BX + C respectively, get: − 1 + B + C = 0C = 5, solve this equation, get: B = − 4C = 5; therefore, the analytical formula of parabola is y = - x2-4x + 5; (2) from y = - x2-4x + 5, let y = 0, then the equation can be solved, We get - x2-4x + 5 = 0, and solve this equation, we get X1 = - 5, X2 = 1; so the coordinate of point C is (- 5, 0). We get point d (- 2, 9) from the vertex coordinate formula. If the perpendicular of X axis intersects with M through D, then s △ DMC = 12 × 9 × (5 − 2) = 272, s trapezoid mdbo = 12 × 2 × (9 + 5) = 14, s △ BOC = 12 × 5 × 5 = 252, so s △ BCD = s trapezoid mdbo + s △ DMC − s △ BOC = 14 + 272 − 252 = 15
It is known that m and N are the two real roots of the equation x ^ 2-6x + 5, and m
1) We can get x = 1 or x = 5 from x ^ 2-6x + 5 = (x-1) (X-5) = 0, so m = 1, n = 5, substituting x = 1, y = 0 and x = 0, y = 5 into the parabolic equation, we can get - 1 + B + C = 0, and C = 5, we can get b = - 4, C = 5, so the analytical formula of the parabolic equation is y = - x ^ 2-4x + 5.2) let y = - x ^ 2-4x + 5 = 0, we can get x = 1
1. Known P: | 1 - (x-1) / 3|
p: X > = - 2 is not p XM or X-1
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