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FP1 2023 Jan Q5

A Level / Edexcel / FP1

IAL 2023 Jan Paper · Question 5

题目

Problem

The quadratic equation

4x2+3x+k=04x^2 + 3x + k = 0

where kk is an integer, has roots α\alpha and β\beta

(a) Write down, in terms of kk where appropriate, the value of α+β\alpha + \beta and the value of αβ\alpha\beta

(2)

(b) Determine, in simplest form in terms of kk, the value of αβ2+βα2\dfrac{\alpha}{\beta^2} + \dfrac{\beta}{\alpha^2}

(4)

(c) Determine a quadratic equation which has roots

αβ2\frac{\alpha}{\beta^2}

and

βα2\frac{\beta}{\alpha^2}

giving your answer in the form px2+qx+r=0px^2 + qx + r = 0 where pp, qq and rr are integer values in terms of kk

(3)
题目中文翻译

二次方程

4x2+3x+k=04x^2 + 3x + k = 0

其中 kk 为整数，两根为 α\alpha 和 β\beta。

(a) 直接写出（用含 kk 的式子表示，若适用）α+β\alpha + \beta 和 αβ\alpha\beta 的值。

(2)

(b) 用含 kk 的最简形式确定 αβ2+βα2\dfrac{\alpha}{\beta^2} + \dfrac{\beta}{\alpha^2} 的值。

(4)

(c) 求一个二次方程，使其两根为

αβ2\frac{\alpha}{\beta^2}

和

βα2\frac{\beta}{\alpha^2}

答案以 px2+qx+r=0px^2 + qx + r = 0 的形式表示，其中 pp、qq、rr 为用含 kk 的式子表示的整数。

(3)

解答

(a)

解法一

思路

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直接使用二次方程的根与系数关系，分别写出两根之和与两根之积。

答题过程

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By the sum and product of roots,

α+β=−34\boxed{\alpha+\beta=-\frac34}

and

αβ=k4.\boxed{\alpha\beta=\frac k4}.

(b)

解法一

思路

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先通分，把分子化为 α3+β3\alpha^3+\beta^3，分母化为 (αβ)2(\alpha\beta)^2。再用立方和恒等式把分子改写成只含 α+β\alpha+\beta 与 αβ\alpha\beta 的形式，并代入 (a)。

答题过程

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Since the given expressions are defined, αβ≠0\alpha\beta\ne0, so k≠0k\ne0. Combining the fractions,

 αβ2+βα2= α⋅α2α2β2 +β⋅β2α2β2= α3+β3α2β2.\begin{align*} &\,\frac{\alpha}{\beta^2}+\frac{\beta}{\alpha^2}\\ =&\,\frac{\alpha\cdot\alpha^2}{\alpha^2\beta^2}\\ &\,\hspace{2pt}+\frac{\beta\cdot\beta^2}{\alpha^2\beta^2}\\ =&\,\frac{\alpha^3+\beta^3}{\alpha^2\beta^2}. \end{align*}

Let N=α3+β3N=\alpha^3+\beta^3. Using the sum of cubes identity,

N= (α+β)3 −3αβ(α+β).\begin{align*} N=&\,(\alpha+\beta)^3\\ &\,\hspace{2pt}-3\alpha\beta(\alpha+\beta). \end{align*}

Using part (a),

N= (−34)3 −3(k4)(−34)= −2764+9k16= −27+36k64= 36k−2764.\begin{align*} N=&\,\Big(-\frac34\Big)^3\\ &\,\hspace{2pt}-3\Big(\frac k4\Big)\Big(-\frac34\Big)\\ =&\,-\frac{27}{64}+\frac{9k}{16}\\ =&\,\frac{-27+36k}{64}\\ =&\,\frac{36k-27}{64}. \end{align*}

The denominator is

D= α2β2= (αβ)2= (k4)2= k216.\begin{align*} D=&\,\alpha^2\beta^2\\ =&\,(\alpha\beta)^2\\ =&\,\Big(\frac k4\Big)^2\\ =&\,\frac{k^2}{16}. \end{align*}

Therefore,

 αβ2+βα2= ND= (36k−27)/64(k/4)2= 36k−2764 ×16k2= 36k−274k2.\begin{align*} &\,\frac{\alpha}{\beta^2}+\frac{\beta}{\alpha^2}\\ =&\,\frac ND\\ =&\,\frac{(36k-27)/64}{(k/4)^2}\\ =&\,\frac{36k-27}{64}\\ &\,\hspace{2pt}\times\frac{16}{k^2}\\ =&\,\boxed{\frac{36k-27}{4k^2}}. \end{align*}

(c)

解法一

思路

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新方程的根和就是 (b) 的结果。另求两个新根的乘积，再代入“x2−x^2- 根和 ×x+\times x+ 根积 =0=0”，最后整体乘以 4k24k^2 得到整数形式的系数。

答题过程

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The product of the new roots is

 (αβ2)(βα2)= αβα2β2= 1αβ= 4k.\begin{align*} &\,\Big(\frac{\alpha}{\beta^2}\Big)\Big(\frac{\beta}{\alpha^2}\Big)\\ =&\,\frac{\alpha\beta}{\alpha^2\beta^2}\\ =&\,\frac1{\alpha\beta}\\ =&\,\frac4k. \end{align*}

Hence the required equation is

 x2−36k−274k2x +4k= 0.\begin{align*} &\,x^2-\frac{36k-27}{4k^2}x\\ &\,\hspace{2pt}+\frac4k\\ =&\,0. \end{align*}

Multiplying by 4k24k^2 gives

 4k2x2 −4k236k−274k2x +4k24k= 0.\begin{align*} &\,4k^2x^2\\ &\,\hspace{2pt}-4k^2\frac{36k-27}{4k^2}x\\ &\,\hspace{4pt}+4k^2\frac4k\\ =&\,0. \end{align*}

Thus the required quadratic equation is

 4k2x2−(36k−27)x +16k= 0.\begin{align*} &\,4k^2x^2-(36k-27)x\\ &\,\hspace{2pt}+16k\\ =&\,0. \end{align*}