4 problems found
Write down the cubes of the integers \(1, 2, \ldots , 10\). The positive integers \(x\), \(y\) and \(z\), where \(x < y\), satisfy \[ x^3+y^3 = kz^3\,, \tag{\(*\)} \] where \(k\) is a given positive integer.
Solution: \begin{array}{c|c} n & n^3 \\ \hline 1 & 1 \\ 2 & 8 \\ 3 & 27 \\ 4 & 64 \\ 5 & 125 \\ 6 & 216 \\ 7 & 343 \\ 8 & 512 \\ 9 & 729 \\ 10 & 1000 \\ \end{array}
Show that \(x^3-3xbc + b^3 + c^3\) can be written in the form \(\left( x+ b+ c \right) {\rm Q}( x)\), where \({\rm Q}( x )\) is a quadratic expression. Show that \(2{\rm Q }( x )\) can be written as the sum of three expressions, each of which is a perfect square. It is given that the equations \(ay^2 + by + c =0\) and \(by^2 + cy + a = 0\) have a common root \(k\). The coefficients \(a\), \(b\) and \(c\) are real, \(a\) and \(b\) are both non-zero, and \(ac \neq b^2\). Show that \[ \left( ac - b^2 \right) k = bc - a^2 \] and determine a similar expression involving \(k^2\). Hence show that \[ \left( ac - b^2 \right) \left(ab-c^2 \right) = \left( bc - a^2 \right)^2 \] and that \( a^3 -3abc + b^3 +c^3 = 0\,\). Deduce that either \(k=1\) or the two equations are identical.
Solution: \begin{align*} && x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\ &&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\ \end{align*} We must have: \begin{align*} && 0 &= ak^2 + bk+c \tag{1}\\ &&0 &= bk^2+ck+a \tag{2}\\ b*(1)&& 0 &= abk^2 + b^2k+cb \\ a*(2)&& 0 &= abk^2 + ack + a^2 \\ \Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\ \Rightarrow && (ac-b^2)k &= bc-a^2 \\ \\ c*(1) && 0 &= ack^2+bck+c^2 \\ b*(2) && 0 &= b^2k^2+bck+ab \\ \Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\ \Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\ \\ \Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\ \Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\ \Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\ \Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\ &&&= a(a^3+b^3+c^3-3abc) \\ \underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\ &&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align*} Therefore \(a+b+c = 0\). (Since otherwise \(a=b=c\) but \(ac \neq b^2\)). This means \(1\) is a root of our equations. Therefore, either \(k = 1\) or they have both roots in common, ie they are the same equation up to a scalar factor. ie \(b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1\). Therefore, they are the same equation.
Give a condition that must be satisfied by \(p\), \(q\) and \(r\) for it to be possible to write the quadratic polynomial \(px^2 + qx + r\) in the form \(p \l x + h \r^2\), for some \(h\). Obtain an equation, which you need not simplify, that must be satisfied by \(t\) if it is possible to write \[ \l x^2 + \textstyle{{1 \over 2}} bx + t \r^2 - \l x^4 + bx^3 + cx^2 +dx +e \r \] in the form \(k \l x + h \r^2\), for some \(k\) and \(h\). Hence, or otherwise, write \(x^4 + 6x^3 + 9x^2 -2x -7\) as a product of two quadratic factors.
Let \(\mathrm{h}(x)=ax^{2}+bx+c,\) where \(a,b\) and \(c\) are constants, and \(a\neq0\). Give a condition which \(a,b\) and \(c\) must satisfy in order that \(\mathrm{h}(x)\) can be written in the form \[ a(x+k)^{2},\tag{*} \] where \(k\) is a constant. If \(\mathrm{f}(x)=3x^{2}+4x\) and \(\mathrm{g}(x)=x^{2}-2\), find the two constant values of \(\lambda\) such that \(\mathrm{f}(x)+\lambda\mathrm{g}(x)\) can be written in the form \((*)\). Hence, or otherwise, find constants \(A,B,C,D,m\) and \(n\) such that \begin{alignat*}{1} \mathrm{f}(x) & =A(x+m)^{2}+B(x+n)^{2}\\ \mathrm{g}(x) & =C(x+m)^{2}+D(x+n)^{2}. \end{alignat*} If \(\mathrm{f}(x)=3x^{2}+4x\) and \(\mathrm{g}(x)=x^{2}+\alpha\) and it is given by that there is only one value of \(\lambda\) for which \(\mathrm{f}(x)+\lambda\mathrm{g}(x)\) can be written in the form \((*)\), find \(\alpha\).
Solution: For \(h(x)\) to be written in this form \(b^2=4ac\). Suppose \(f(x) = 3x^2+4x\), \(g(x) = x^2-2\). then, \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x - 2 \lambda \\ \Rightarrow && 0 &= 16 + 8(3+\lambda) \lambda \\ \Rightarrow && 0 &= 2+ 3 \lambda + \lambda^2 \\ &&&= (\lambda +1)(\lambda + 2) \\ \Rightarrow && \lambda &= -1 , -2 \\ \end{align*} \begin{align*} && f(x) - g(x) &= 2(x+1)^2 \\ && f(x) -2g(x) &= (x+2)^2 \\ \Rightarrow && g(x) &= 2(x+1)^2 - (x+2)^2 \\ && f(x) &= 4(x+1)^2 - (x+2)^2 \end{align*} Suppose \(f(x) = 3x^2+4x, g(x) = x^2 + \alpha\), then \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x+\lambda \alpha \\ \Rightarrow && 0 &= 16 -2\lambda \alpha(\lambda + 3) \\ && 0 &= \alpha \lambda^2 +3\lambda-8 \\ \Rightarrow && 0 &= 9 +32 \alpha \\ \Rightarrow && \alpha &= -\frac{9}{32} \end{align*}