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第四章:部分分式 Partial Fraction

第六章:角的形成及单位 Angles and Measurements

第十三章: 方程组 Simultaneous Equations

第十五章:二元一次不等式及线性规划 linear inequality in two variables and linear programming

2.2 一元二次方程式的根的判别式 The Discriminant of Quadratic Equations

2.2 The Discriminant of Quadratic Equations 一元二次方程式的根的判别式


 

[1]

The quadratic equation kx^2 + 2(k + a)x + (k + b) = 0 has equal roots. Express k in terms of a and b.

一元二次方程式kx^2 + 2(k + a)x + (k + b) = 0 有等根. 用a 和 b 表示k.

Answer:
k = \frac{a^2}{b-2a}
[2]

Find the range of k where the equation kx^2+ x + 2 = (x + 1)^2 has two unequal real roots.

求k 的范围当方程式 kx^2+ x + 2 = (x + 1)^2 有两个相异的实根。

Answer:
k < 1\frac{1}{4}
[3]

The quadratic equation (p + 1)x^2 + 2px + (p + 2) = 0 has real roots. Find the range of values of p.

一元二次方程式(p + 1)x^2 + 2px + (p + 2) = 0 有实根. 求 p值的范围.

Answer:
p \leq -\frac{2}{3}

[4]

Prove that 2ax^2 + (3a + 2c)x + 2c = 0 has distinct real root for all real value of x.

试证对所有的实数x,2ax^2 + (3a + 2c)x + 2c = 0 都有相异的实根.


 

[5]

Find the value of k so that the line kx + y = 8 is the tangent of the curve x^2 + xy = 4.

求k 的值使得直线 kx + y = 8 是曲线 x^2 + xy = 4的切线.

Answer:
5
[6]

Prove that, for all real value of a, the equation x^2 + 2ax + 2a^2 + a + 1 = 0 has no real roots for x.

试证对所有的实数a, 方程式 x^2 + 2ax + 2a^2 + a + 1 = 0 的x都没有实根.


 

[7]

7Show that the line x + y = q will intersect the curve x^2 – 2x + 2y^2 = 3 in two distinct point if q^2 < 2q + 5. 试证若q^2 < 2q + 5,直线x + y = q 与曲线x^2 – 2x + 2y^2 = 3 交于两个相异的点。


 

[8]

Find the range of the values of c for which 3x^2 + 5x +c is always positive.

求c 值的范围使得 3x^2 + 5x +c 恒为正。

Answer:
c > \frac{25}{12}
[9]

If the quadratic function f (x) = (a – 2)x^2 – 2ax + a + 1 is never positive, find the greatest value of a.

若f (x) = (a – 2)x^2 – 2ax + a + 1 是非正二元一次函数, 求 a的最大值.

Answer:
-2
[10]

Find the values of a for which ax^2 + 3x + 4a is
求a 值使得 ax^2 + 3x + 4a 为

(a) positive for all the real values of x,
正对所有的实数x,

(b) negative for all the real values of x.
负对所有的实数x

Answer:
(a) a > \frac{3}{4}

(b) a < -\frac{3}{4}

[11]

Prove that (mx – 1)(x – 2) = m has two distinct real roots for x \epsilon R.

试证在x \epsilon R,(mx – 1)(x – 2) = m 有两个相异的实根。