CalcGospel 國際數學圖譜
返回

FP1 2022 June Q5

A Level / Edexcel / FP1

IAL 2022 June Paper · Question 5

题目

Problem

5. The quadratic equation

2x2−3x+5=02x^2 - 3x + 5 = 0

has roots α\alpha and β\beta

Without solving the equation,

(a) write down the value of (α+β)(\alpha + \beta) and the value of αβ\alpha\beta

(1)

(b) determine the value of

(i) α2+β2\alpha^2 + \beta^2

(ii) α3+β3\alpha^3 + \beta^3

(4)

(c) find a quadratic equation which has roots

(α3−β)(\alpha^3-\beta)

and

(β3−α)(\beta^3-\alpha)

giving your answer in the form px2+qx+r=0px^2 + qx + r = 0 where pp, qq and rr are integers to be determined.

(5)
题目中文翻译
  1. 二次方程
2x2−3x+5=02x^2 - 3x + 5 = 0

的根为 α\alpha 和 β\beta。

不解方程，

(a) 写出 (α+β)(\alpha + \beta) 和 αβ\alpha\beta 的值。

(1)

(b) 求下列各式的值

(i) α2+β2\alpha^2 + \beta^2

(ii) α3+β3\alpha^3 + \beta^3

(4)

(c) 求一个二次方程，使其根为

(α3−β)(\alpha^3-\beta)

和

(β3−α)(\beta^3-\alpha)

答案写成 px2+qx+r=0px^2 + qx + r = 0 的形式，其中 pp、qq 和 rr 是待定整数。

(5)

解答

(a)

解法一

思路

展开

使用二次方程的根与系数关系：根的和为一次项系数的相反数除以二次项系数，根的积为常数项除以二次项系数。

答题过程

展开

For 2x2−3x+5=02x^2-3x+5=0,

α+β= 32,αβ= 52.\begin{align*} \alpha+\beta=&\,\boxed{\frac32},\\ \alpha\beta=&\,\boxed{\frac52}. \end{align*}

(b)(i)

解法一

思路

展开

将平方和写成 (α+β)2−2αβ(\alpha+\beta)^2-2\alpha\beta，再代入 (a) 的结果。

答题过程

展开 α2+β2= (α+β)2−2αβ= (32)2 −2(52)= 94−5= −114.\begin{align*} \alpha^2+\beta^2 =&\,(\alpha+\beta)^2-2\alpha\beta\\ =&\,\left(\frac32\right)^2\\ &\,\hspace{2pt}-2\left(\frac52\right)\\ =&\,\frac94-5\\ =&\,\boxed{-\frac{11}{4}}. \end{align*}

(b)(ii)

解法一

思路

展开

由立方和恒等式

 α3+β3= (α+β)3 −3αβ(α+β),\begin{align*} &\,\alpha^3+\beta^3\\ =&\,(\alpha+\beta)^3\\ &\,\hspace{2pt}-3\alpha\beta(\alpha+\beta), \end{align*}

直接使用 (a) 的根和与根积。

答题过程

展开  α3+β3= (α+β)3 −3αβ(α+β)= (32)3 −3(52)(32)= 278−908= −638.\begin{align*} &\,\alpha^3+\beta^3\\ =&\,(\alpha+\beta)^3\\ &\,\hspace{2pt}-3\alpha\beta(\alpha+\beta)\\ =&\,\left(\frac32\right)^3\\ &\,\hspace{2pt}-3\left(\frac52\right) \left(\frac32\right)\\ =&\,\frac{27}{8}-\frac{90}{8}\\ =&\,\boxed{-\frac{63}{8}}. \end{align*}

(c)

解法一

思路

展开

设新方程的两根为 u=α3−βu=\alpha^3-\beta、v=β3−αv=\beta^3-\alpha。先利用 (a)、(b) 求 u+vu+v；求 uvuv 时展开，并用

 α4+β4= (α2+β2)2 −2(αβ)2\begin{align*} &\,\alpha^4+\beta^4\\ =&\,(\alpha^2+\beta^2)^2\\ &\,\hspace{2pt}-2(\alpha\beta)^2 \end{align*}

化简。最后按新两根的和与积建立方程：

 x2−(u+v)x +uv=0,\begin{align*} &\,x^2-(u+v)x\\ &\,\hspace{2pt}+uv=0, \end{align*}

再清除分母。

答题过程

展开

Let

u= α3−β,v= β3−α.\begin{align*} u=&\,\alpha^3-\beta,\\ v=&\,\beta^3-\alpha. \end{align*}

Using parts (a) and (b),

u+v= α3+β3−(α+β)= −638−32= −758.\begin{align*} u+v =&\,\alpha^3+\beta^3-(\alpha+\beta)\\ =&\,-\frac{63}{8}-\frac32\\ =&\,-\frac{75}{8}. \end{align*}

Also,

 α4+β4= (α2+β2)2 −2(αβ)2= (−114)2 −2(52)2= 12116−20016= −7916.\begin{align*} &\,\alpha^4+\beta^4\\ =&\,(\alpha^2+\beta^2)^2\\ &\,\hspace{2pt}-2(\alpha\beta)^2\\ =&\,\left(-\frac{11}{4}\right)^2\\ &\,\hspace{2pt}-2\left(\frac52\right)^2\\ =&\,\frac{121}{16}-\frac{200}{16}\\ =&\,-\frac{79}{16}. \end{align*}

Therefore,

 uv= (α3−β)(β3−α)= (αβ)3 −(α4+β4)+αβ= (52)3 −(−7916)+52= 250+79+4016= 36916.\begin{align*} &\,uv\\ =&\,(\alpha^3-\beta)(\beta^3-\alpha)\\ =&\,(\alpha\beta)^3\\ &\,\hspace{2pt}-(\alpha^4+\beta^4)+\alpha\beta\\ =&\,\left(\frac52\right)^3\\ &\,\hspace{2pt}-\left(-\frac{79}{16}\right) +\frac52\\ =&\,\frac{250+79+40}{16}\\ =&\,\frac{369}{16}. \end{align*}

Hence the required quadratic equation is

x2−(u+v)x+uv= 0,x2+758x+36916= 0.\begin{align*} x^2-(u+v)x+uv=&\,0,\\ x^2+\frac{75}{8}x+\frac{369}{16}=&\,0. \end{align*}

Multiplying by 1616 gives

16x2+150x+369=0.\boxed{16x^2+150x+369=0}.

解法二

思路

展开

将根和的四次方展开，再移去混合项，得到四次和。这样只需要 (a)、(b)(i) 中的根和、根积与平方和，便可求出新两根的积。

答题过程

展开

Let u=α3−βu=\alpha^3-\beta and v=β3−αv=\beta^3-\alpha. As in the first method,

u+v=−758.u+v=-\frac{75}{8}.

Expanding the fourth power gives

 (α+β)4= α4+β4 +4αβ(α2+β2) +6α2β2.\begin{align*} &\,(\alpha+\beta)^4\\ =&\,\alpha^4+\beta^4\\ &\,\hspace{2pt}+4\alpha\beta(\alpha^2+\beta^2)\\ &\,\hspace{4pt}+6\alpha^2\beta^2. \end{align*}

Hence

 α4+β4= (α+β)4 −4αβ(α2+β2) −6(αβ)2= (32)4 −4(52)(−114) −6(52)2= 8116+44016−60016= −7916.\begin{align*} &\,\alpha^4+\beta^4\\ =&\,(\alpha+\beta)^4\\ &\,\hspace{2pt}-4\alpha\beta(\alpha^2+\beta^2)\\ &\,\hspace{4pt}-6(\alpha\beta)^2\\ =&\,\bigg(\frac32\bigg)^4\\ &\,\hspace{2pt}-4\bigg(\frac52\bigg)\bigg(-\frac{11}{4}\bigg)\\ &\,\hspace{4pt}-6\bigg(\frac52\bigg)^2\\ =&\,\frac{81}{16}+\frac{440}{16}-\frac{600}{16}\\ =&\,-\frac{79}{16}. \end{align*}

Expanding the product of the new roots,

 uv= (α3−β)(β3−α)= (αβ)3 −(α4+β4)+αβ= (52)3 −(−7916)+52= 250+79+4016= 36916.\begin{align*} &\,uv\\ =&\,(\alpha^3-\beta)(\beta^3-\alpha)\\ =&\,(\alpha\beta)^3\\ &\,\hspace{2pt}-(\alpha^4+\beta^4)+\alpha\beta\\ =&\,\bigg(\frac52\bigg)^3\\ &\,\hspace{2pt}-\bigg(-\frac{79}{16}\bigg)+\frac52\\ =&\,\frac{250+79+40}{16}\\ =&\,\frac{369}{16}. \end{align*}

Therefore the required equation is

x2+758x+36916=0.x^2+\frac{75}{8}x+\frac{369}{16}=0.

Multiplying by 1616 gives

16x2+150x+369=0.\boxed{16x^2+150x+369=0}.

解法三

思路

展开

把立方和与根和相乘，展开后除了四次和，剩下的项可用根积乘平方和表示。这样可以直接承接 (b) 的两个结果。

答题过程

展开

Again let u=α3−βu=\alpha^3-\beta and v=β3−αv=\beta^3-\alpha, so that

u+v=−758.u+v=-\frac{75}{8}.

Now

 (α3+β3)(α+β)= α4+α3β +αβ3+β4= α4+β4 +αβ(α2+β2).\begin{align*} &\,(\alpha^3+\beta^3)(\alpha+\beta)\\ =&\,\alpha^4+\alpha^3\beta\\ &\,\hspace{2pt}+\alpha\beta^3+\beta^4\\ =&\,\alpha^4+\beta^4\\ &\,\hspace{2pt}+\alpha\beta(\alpha^2+\beta^2). \end{align*}

Rearranging and using parts (a) and (b),

 α4+β4= (α3+β3)(α+β) −αβ(α2+β2)= (−638)(32) −(52)(−114)= −18916+11016= −7916.\begin{align*} &\,\alpha^4+\beta^4\\ =&\,(\alpha^3+\beta^3)(\alpha+\beta)\\ &\,\hspace{2pt}-\alpha\beta(\alpha^2+\beta^2)\\ =&\,\bigg(-\frac{63}{8}\bigg)\bigg(\frac32\bigg)\\ &\,\hspace{2pt}-\bigg(\frac52\bigg)\bigg(-\frac{11}{4}\bigg)\\ =&\,-\frac{189}{16}+\frac{110}{16}\\ =&\,-\frac{79}{16}. \end{align*}

Therefore,

 uv= (αβ)3 −(α4+β4)+αβ= (52)3 −(−7916)+52= 36916.\begin{align*} &\,uv\\ =&\,(\alpha\beta)^3\\ &\,\hspace{2pt}-(\alpha^4+\beta^4)+\alpha\beta\\ =&\,\bigg(\frac52\bigg)^3\\ &\,\hspace{2pt}-\bigg(-\frac{79}{16}\bigg)+\frac52\\ =&\,\frac{369}{16}. \end{align*}

Using the sum and product of u,vu,v, the quadratic equation is

x2+758x+36916=0,x^2+\frac{75}{8}x+\frac{369}{16}=0,

or, with integer coefficients,

16x2+150x+369=0.\boxed{16x^2+150x+369=0}.