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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

2.3 一元二次方程式的根与系数的关系 The Relations of Roots and Coefficients of Quadratic Equations

2.3 The Relations of Roots and Coefficients of Quadratic Equations 一元二次方程式的根与系数的关系


 

[1]

If \alpha, \beta are the roots of 3x^2 – 2x – 3 = 0, find the values of

\alpha, \beta 是 3x^2 – 2x – 3 = 0 的根, 求值

[a] \frac{1}{\alpha}+\frac{1}{\beta}

[b] \alpha^2 + \beta^2

[c] \frac{\beta}{\alpha}+\frac{\alpha}{\beta}

Answer:
(a) \frac{2}{3}

(b) 2\frac{4}{9}

(c) -2\frac{4}{9}

[2]

The equation 3x^2 + 10x + p = 0, has roots \alpha and \beta where \alpha\beta = 2\frac{2}{3} , find the value of p.

方程式3x^2 + 10x + p = 0 有根 \alpha\beta 其中 \alpha\beta = 2\frac{2}{3} , 求 p的值.

Answer:
3
[3]

The equation 2x^2 – 2x + 3 = 0 has roots p and q and the equation x^2 – x + 2m = 0 has roots and . Find the values of k and m.

方程式2x^2 – 2x + 3 = 0 有根 p 和 q 及方程式 x^2 – x + 2m = 0 有根 和 . 求 k 和 m的值.

Answer:
1\frac{1}{2} , \frac{3}{4}
[4]

If the square of the difference of the roots of equation x^2 – mx +15 = 0 is 4, find the value of m.

若方程式x^2 – mx +15 = 0 的根的平方差是 4, 求 m 的值.

Answer:
\pm 8
[5]

If \alpha and \beta are the roots of equation 2x^2– 3x+ 4 = 0, find the equation whose roots are

\alpha\beta 是方程式2x^2– 3x+ 4 = 0 的根, 求作方程式其根是

[a] \alpha+\frac{1}{\alpha} , \beta+\frac{1}{\beta}

[b] \alpha^2 , \beta^2

Answer:
(a) 8x^2 – 18x + 13 = 0

(b) 4x^2 + 7x + 16 = 0

[6]

One root of the equation 2x^2 – x + c = 0 is twice the other. Find the value of c.

方程式2x^2 – x + c = 0 其中一根是另外一根的两倍. 求 c的值.

Answer:
\frac{1}{9}
[7]

One root of the equation 3x^2 – 3px + p^2 = p + 6 is twice of the other. Find the values of p.

方程式3x^2 – 3px + p^2 = p + 6 其中一根是另外一根的一半. 求 p的值.

Answer:
6, -3
[8]

Given that \alpha and \beta are the roots of the equation 2x^2 = 3x – 4,

已知 \alpha\beta 是方程式 2x^2 = 3x – 4的根,

[a]

form an equation whose roots are \alpha\beta and \beta\alpha.

作方程式其根是\alpha\beta and \beta\alpha.

[b]

show that 4\alpha^3 = \alpha – 12.

证明4\alpha^3 = \alpha – 12.

Answer:
(a) 4x^2 + 23 = 0