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FP1 2021 Oct Q3

A Level / Edexcel / FP1

IAL 2021 Oct Paper · Question 3

题目

Problem

3. The quadratic equation

2x2−5x+7=02x^2 - 5x + 7 = 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, giving each answer as a simplified fraction, the value of

(i) α2+β2\alpha^2 + \beta^2

(ii) α3+β3\alpha^3 + \beta^3

(4)

(c) find a quadratic equation that has roots

 1α+β2 and 1β+α2\begin{align*} &\,\frac{1}{\alpha+\beta^2}\\ &\,\text{and }\frac{1}{\beta+\alpha^2} \end{align*}

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

(4)
题目中文翻译
  1. 二次方程
2x2−5x+7=02x^2 - 5x + 7 = 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（答案用最简分数表示）：

(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

 1α+β2 和 1β+α2\begin{align*} &\,\frac{1}{\alpha+\beta^2}\\ &\,\text{和 }\frac{1}{\beta+\alpha^2} \end{align*}

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。

(4)

解答

(a)

解法一

思路

展开

直接使用二次方程的根与系数关系：两根之和为一次项系数的相反数除以二次项系数，两根之积为常数项除以二次项系数。

答题过程

展开

For the equation 2x2−5x+7=02x^2-5x+7=0, the sum and product of the roots are

α+β=−−52=52\alpha+\beta=-\frac{-5}{2}=\boxed{\frac52}

and

αβ=72.\alpha\beta=\boxed{\frac72}.

(b)(i)

解法一

思路

展开

利用恒等式

 α2+β2= (α+β)2−2αβ,\begin{align*} &\,\alpha^2+\beta^2\\ =&\,(\alpha+\beta)^2-2\alpha\beta, \end{align*}

代入 (a) 的结果。

答题过程

展开  α2+β2= (α+β)2−2αβ= (52)2−2(72)= 254−7= 25−284= −34.\begin{align*} &\,\alpha^2+\beta^2\\ =&\,(\alpha+\beta)^2-2\alpha\beta\\ =&\,\big(\tfrac52\big)^2-2\big(\tfrac72\big)\\ =&\,\frac{25}{4}-7\\ =&\,\frac{25-28}{4}\\ =&\,\boxed{-\frac34}. \end{align*}

(b)(ii)

解法一

思路

展开

利用立方和恒等式，把 α3+β3\alpha^3+\beta^3 改写成只含 α+β\alpha+\beta 与 αβ\alpha\beta 的式子。

答题过程

展开  α3+β3= (α+β)3 −3αβ(α+β)= (52)3 −3(72)(52)= 1258−1054= 125−2108= −858.\begin{align*} &\,\alpha^3+\beta^3\\ =&\,(\alpha+\beta)^3\\ &\,-3\alpha\beta(\alpha+\beta)\\ =&\,\big(\tfrac52\big)^3\\ &\,-3\big(\tfrac72\big)\big(\tfrac52\big)\\ =&\,\frac{125}{8}-\frac{105}{4}\\ =&\,\frac{125-210}{8}\\ =&\,\boxed{-\frac{85}{8}}. \end{align*}

(c)

解法一

思路

展开

把题目给出的两个新根记为 uu、vv。分别求 u+vu+v 与 uvuv，其中分母展开后可用 (a)、(b) 的对称式结果化简；最后使用根为 uu、vv 的二次方程

x2−(u+v)x+uv=0.x^2-(u+v)x+uv=0.

答题过程

展开

Let

u=1α+β2u=\frac{1}{\alpha+\beta^2}

and

v=1β+α2.v=\frac{1}{\beta+\alpha^2}.

The common denominator required for both the sum and the product is

D= (α+β2)(β+α2)= αβ+α3+β3 +α2β2= αβ+(α3+β3) +(αβ)2= 72−858+(72)2= 72−858+494= 28−85+988= 418.\begin{align*} D =&\,(\alpha+\beta^2)(\beta+\alpha^2)\\ =&\,\alpha\beta+\alpha^3+\beta^3\\ &\,+\alpha^2\beta^2\\ =&\,\alpha\beta+(\alpha^3+\beta^3)\\ &\,+(\alpha\beta)^2\\ =&\,\frac72-\frac{85}{8}+\big(\tfrac72\big)^2\\ =&\,\frac72-\frac{85}{8}+\frac{49}{4}\\ =&\,\frac{28-85+98}{8}\\ =&\,\frac{41}{8}. \end{align*}

Since D≠0D\ne0, both new roots are defined. Their sum is

 u+v= β+α2+α+β2D= α2+β2+α+βD= −34+52418= 74418= 74×841= 1441.\begin{align*} &\,u+v\\ =&\,\frac{\beta+\alpha^2+\alpha+\beta^2}{D}\\ =&\,\frac{\alpha^2+\beta^2+\alpha+\beta}{D}\\ =&\,\frac{-\frac34+\frac52}{\frac{41}{8}}\\ =&\,\frac{\frac74}{\frac{41}{8}}\\ =&\,\frac74\times\frac{8}{41}\\ =&\,\frac{14}{41}. \end{align*}

Their product is

uv= 1D= 1418= 841.\begin{align*} uv=&\,\frac{1}{D}\\ =&\,\frac{1}{\frac{41}{8}}\\ =&\,\frac{8}{41}. \end{align*}

Hence the required quadratic equation is

x2−(u+v)x+uv=0,x^2-(u+v)x+uv=0,

that is,

x2−1441x+841=0.x^2-\frac{14}{41}x+\frac{8}{41}=0.

Multiplying by 4141 gives

41x2−14x+8=0.41x^2-14x+8=0.

Therefore,

41x2−14x+8=0.\boxed{41x^2-14x+8=0}.