Quadratic equation again!!

Angela

2010-09-29 6:14 am

If the quadratic equation 6x^2-7X-1/3=2k has 2 distinct real roots,
i) Find the range of possible values of k
Ans: k>-19/16
ii)find the roots of the equation when k is the negative integer in (i)
Ans: x=5/6 or x=1/3
b) Using the result of (ii), solve -3(x+1)^2+7/2(x+1)-5/6=0
I just want to know the answer of part b, I don't understand this question *o*

回答 (1)

用戶2053

2010-09-29 6:31 am

✔ 最佳答案

i) Rewrite the equation as 6x^2 - 7X - (1/3 + 2k) = 0△ > 07^2 - 4 * 6 * [- (1/3 + 2k)] > 049 + 24(1/3 + 2k) > 048k + 8 + 49 > 048k > - 57k > - 19/16ii) k is the negative integer which is > - 19/16 = - 1.1875So k = - 1 ,The equation becomes 6x^2 - 7x - (1/3 + 2(-1)) = 018x^2 - 21x - (1 - 6) = 018x^2 - 21x + 5 = 0 ...........(*)(3x - 1)(6x - 5) = 0x = 1/3 or x = 5/6b) 18x^2 - 21x + 5 = 0 ...........(*)(- 1/6) (18x^2 - 21x + 5) = 0(- 18/6)x^2 + (21/6)x - 5/6 = 0- 3x^2 + (7/2)x - 5/6 = 0 ...(A)We know the root of (A) are x = 1/3 or x = 5/6 ,So the roots of -3(x+1)^2 + (7/2) (x+1) - 5/6=0 are x = 1/3 - 1 or x = 5/6 - 1 ,So x = - 2/3 or x = - 1/6

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