F4 Qadratic equations 問題

Vicky

2009-11-05 6:34 am

請問以下呢幾題係點計架~
請各位幫幫手 plz~

1. if A and B are the roots of the Quadratic Equation x(2x-7)+1=0, form a quadraic equation in x whose roots are the negatives of the reciprocals ofA and B.

2. it is given that the graph of y=(m-2)x^2 + (2m+5)x + (m+1) cuts the x-axis at two points.
(a) Determine whether the value of m can be 2. Explain your answer.
(b) Find the range of possible values of m.

3. given the quadratic equations (3p+1)x^2 - (p+3)x +1 =0, find the value(s) of p for each of the following cases.
(a) the product of the roots of the equation is -4.
(b) the sum of the equation is 7.
(c) the equation has a repeated real root.

ans.
1. x^2 +7x +2 =0
2. (a) no
(b) m>-11/8 and m=/=2
3. (a) -5/12
(b) -a/5
(c) 1 or 5

回答 (1)

用戶9112

2009-11-05 7:26 am

✔ 最佳答案

(1) x(2x - 7) + 1 = 0
2x^2 - 7x + 1 = 0
Sum of roots A + B = 7/2
Product of roots AB = 1/2
New sum of roots (-1/A) + (-/B) = -(A + B)/AB = -(7/2) / (1/2) = -7
New product of roots = (-1/A)(-1/B) = 1/AB = 2
New equation is x^2 + 7x + 2 = 0
(2a) m cannot be 2 since the equation will reduce to y = 9x + 3 which is not quadratic and can only has 1 root
(b) For 2 real roots, discriminant > 0
(2m + 5)^2 - 4(m - 2)(m + 1) > 0
4m^2 + 20m + 25 - 4m^2 + 4m + 8 > 0
24m + 33 > 0
24m > -33
m > -11/8 and of course m <> 2
(3) (a) Product of roots = 1/(3p + 1) = -4
1 = -12p - 4
12p = -5
p = -5/12
(b) Sum of roots = (p + 3)/(3p+1) = 7
p + 3 = 21p + 7
-4 = 20p
p = -1/5
(c) For repeated roots, discriminant = 0
(p + 3)^2 - 4(3p+1)(1) = 0
p^2 + 6p + 9 - 12p - 4 = 0
p^2 - 6p + 5 = 0
(p - 5)(p - 1) = 0
p = 1 or p = 5

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