f4 maths quadratic equation

[匿名]

2014-10-10 3:22 am

1.x^2+24x+n^2 is a perfect aquare,find n.
2.nature of the roots of x^2+2x-5=0
3.slove(2x+1)^2=(3x-1)(2x+1)
4.k<0,determine the sign of rpots of x^2-x+6k=0
5.sketch y=x^2+3x-1
6.If one root of x^2-ax-3a=0 is 6,find the another root.
要連埋步驟+講解
thanks

更新1:

q1 should be x^2+24x+n^2 is a perfect square,find n. and q2 should be solve(2x+1)^2=(3x-1)(2x+1)

回答 (1)

用戶10012

2014-10-10 3:13 pm

✔ 最佳答案

1.
x^2 + 24x + n^2 = [(x + 12)^2 - 144] + n^2, for this to be a perfect square,
n^2 - 144 = 0, so n = +/- sqrt 144 = +/- 12.
2/3.
(2x + 1)^2 = (3x - 1)(2x + 1)
(2x + 1)^2 - (3x - 1)(2x + 1) = 0
(2x + 1)[(2x + 1) - (3x - 1)] = 0
(2x + 1)(- x + 2) = 0
x = - 1/2 or 2.
4.
Product of roots = c/a = 6k. Since k is < 0, c/a < 0, that is product of roots must be negative, so the 2 roots must be of opposite signs.
6.
Since 6 is one of the roots, sub it into the equation, 6^2 - 6a - 3a = 0
36 - 9a = 0, a = 4.
The equation is x^2 - 4x - 12 = 0, (x - 6)(x + 2) = 0, so the other root is - 2.

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