Differentiation (Part-38) | S N De

   In the previous article , we have discussed few Short Answer Type Questions (From Qst. 41-46) of S.N. Dey Math Exercise. In this article,  we are going to start Long Answer Type Questions of Quadratic Equations (Part-6) as a part of our S.N. Dey Math solution series.  So, let's start.

1. If $\,\,\alpha \pm\sqrt{\beta}\,\,$ be the roots of the equation $\,\,x^2+px+q=0,\,\,$ prove that $\,\,\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}\,\,$ will be the roots of the equation $\,\,(p^2-4q)(p^2x^2+4px)-16q=0=0.$

Sol.  Since $\,\,\alpha \pm\sqrt{\beta}\,\,$ be the roots of the equation $\,\,x^2+px+q=0,\,\,$ we have , sum of the roots $=(\alpha +\sqrt{\beta})+(\alpha -\sqrt{\beta})=-p \\ \Rightarrow  2\alpha=-p \\ \Rightarrow p=-2\alpha \\ \text{and the product of the roots} \\=(\alpha +\sqrt{\beta})(\alpha -\sqrt{\beta})=q \\ \Rightarrow \alpha^2-\beta=q \\ \Rightarrow \frac{p^2}{4}-\beta=q\,\,[\text{Since,}\,\,\alpha=-\frac p2] \\ \Rightarrow p^2-4q=4\beta \\ \text{Now,}\,\,(p^2-4q)(p^2x^2+4px)-16q=0 \\ \Rightarrow 4 \beta(4 \alpha^2x^2 -8 \alpha x)-16(\alpha^2-\beta)=0 \\ \Rightarrow 16 \beta(\alpha^2 x^2-2 \alpha x)-16(\alpha^2-\beta)=0 \\ \Rightarrow \beta(\alpha^2 x^2-2 \alpha x)-\alpha^2+\beta=0 \\ \Rightarrow \beta(\alpha^2 x^2-2 \alpha x+1)=\alpha^2 \\ \Rightarrow (\alpha x-1)^2=\frac{\alpha^2}{\beta} \\ \Rightarrow \alpha x-1=\pm \frac{\alpha}{\sqrt{\beta}} \\ \Rightarrow x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}.$

2.If $\,\,\alpha, \beta\,\,$ and $\,\,\gamma, \delta\,$ be the roots of the equations $\,\,x^2-bx+c=0\,\,$ and $\,\,x^2-cx+b=0\,$ respectively, from the equation whose roots are $\,\,\left(\frac{1}{\alpha \gamma}+\frac{1}{\beta\delta}\right)\,\,$ and $\,\,\left(\frac{1}{\alpha\delta}+\frac{1}{\beta\gamma}\right).$

Sol. Since $\,\,\alpha, \beta\,\,$ be the roots of the equation $\,\,x^2-bx+c=0\,\,$ we have, $\,\, \alpha+\beta=b, \\ \alpha\beta=c.$

Again, since $\,\,\gamma, \delta\,$ be the roots of the equation $\,\,x^2-cx+b=0\,$ we have , $\,\,\gamma+\delta=c,\\ \gamma\delta=b $

Now, the sum of roots of the required equation $=\left(\frac{1}{\alpha \gamma}+\frac{1}{\beta\delta}\right)+\left(\frac{1}{\alpha\delta}+\frac{1}{\beta\gamma}\right)\\=\frac{1}{\gamma}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\frac{1}{\delta}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\\=\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \left(\frac{1}{\gamma}+\frac{1}{\delta}\right)\\=\left(\frac{\alpha+\beta}{\alpha\beta}\right)\left(\frac{\gamma+\delta}{\gamma\delta}\right)\\=\frac bc \times \frac cb\\=1.$

 The product of roots of the required equation $=\left(\frac{1}{\alpha \gamma}+\frac{1}{\beta\delta}\right) \times \left(\frac{1}{\alpha\delta}+\frac{1}{\beta\gamma}\right)\\=\frac{1}{\alpha^2\gamma\delta} \times \frac{1}{\alpha\beta\gamma^2} \times \frac{1}{\alpha \beta \delta^2} \times \frac{1}{\beta^2\gamma\delta}\\=\frac{1}{\gamma\delta} \left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)+\frac{1}{\alpha \beta}\left(\frac{1}{\gamma^2}+\frac{1}{\delta^2}\right)\\=\frac 1b.\left(\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\right)+\frac 1c.\left(\frac{\gamma^2+\delta^2}{\gamma^2\delta^2}\right)\\=\frac{1}{b}. \left(\frac{(\alpha+\beta)^2-2 \alpha\beta}{c^2}\right)+\frac 1c. \left(\frac{(\gamma+\delta)^2-2\gamma\delta}{b^2}\right)\\=\frac{1}{bc}\left(\frac{b^2-2c}{c}+\frac{c^2-2b}{b}\right)\\=\frac{1}{b^2c^2}(b^3+c^3-4bc)$

So, the required quadratic equation is : $\,\,x^2-(\text{the sum of roots})x\\+(\text{ the product of roots})=0 \\ \Rightarrow x^2-x+\frac{1}{b^2c^2}(b^3+c^3-4bc)=0 \\ \Rightarrow b^2c^2x(x-1)+b^3+c^3-4bc=0.$

3.  If $\,\,\alpha,\,\beta\,\,$ be the  roots of the quadratic equation $\,x^2+x+1=0,\,\,$ form the equation whose roots are $\,\,\frac{\alpha}{\beta}\,$ and $\,\,\frac{\beta}{\alpha}.$ Explain why you get the same equation as the given one.

Sol. Since $\,\,\alpha,\,\beta\,\,$ be the  roots of the quadratic equation $\,x^2+x+1=0, \cdots(1) \\ \alpha+\beta=-1,\,\alpha\beta=1.$

The equation of quadratic equation having roots $\,\,\frac{\alpha}{\beta}\,$ and $\,\,\frac{\beta}{\alpha}\,\,$ is given by: $\,\,x^2-\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)x+\frac{\alpha}{\beta} \times \frac{\beta}{\alpha}=0 \\ \Rightarrow x^2-\left(\frac{\alpha^2+\beta^2}{\alpha\beta}\right)x+1=0 \\ \Rightarrow x^2-\left(\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)x+1=0 \\ \Rightarrow x^2-\left(\frac{(-1)^2-2 \times 1}{1}\right)x+1=0 \\ \Rightarrow x^2+x+1=0.$

Now, $\,\,\alpha \,$ being the root of (1), we get $\,\alpha^2+\alpha+1=0 \\ \Rightarrow \alpha^2-\beta=0,\,\,[\text{Since,}\,\,\alpha+\beta=-1]\\ \Rightarrow \alpha=\frac{\beta}{\alpha}\cdots(2)$

Similarly, $\,\,\beta \,$ being the root of (1), we get $\,\beta^2+\beta+1=0 \\ \Rightarrow \beta^2-\alpha=0,\,\,[\text{Since,}\,\,\alpha+\beta=-1]\\ \Rightarrow \beta=\frac{\alpha}{\beta} \cdots(3)$ 

Hence , from (1), (2) and (3), we can conclude that if $\,\,\alpha,\,\beta\,\,$ be the  roots of the quadratic equation $\,x^2+x+1=0,\,\,$ then the equation whose roots are  $\,\,\frac{\alpha}{\beta}\,$ and $\,\,\frac{\beta}{\alpha},$ is the same equation with the aforesaid one.

4. If $\,\,\alpha,\,\beta\,\,$ be the roots of the quadratic equation $\,\,x^2-px+q=0,\,\,$ find the equation whose roots are $\,\,\frac{q}{p-\alpha}\,\,$ and $\,\,\frac{q}{p-\beta}.\,$ Account for the identity of the equation obtained with the given equation.

Sol.  Since $\,\,\alpha,\,\beta\,\,$ be the roots of the quadratic equation $\,\,x^2-px+q=0 \cdots(1),\,\,\\ \alpha+ \beta=p,\,\, \alpha\beta=q.$

Now, the required equation  whose roots are $\,\,\frac{q}{p-\alpha}\,\,$ and $\,\,\frac{q}{p-\beta}\,$ is given by : $\,x^2-\left(\frac{q}{p-\alpha}+\frac{q}{p-\beta}\right)x+\frac{q}{p-\alpha} \times \frac{q}{p-\beta}=0.\cdots(2)$

Next, we calculate $\,\,\left(\frac{q}{p-\alpha}+\frac{q}{p-\beta}\right)\,\, \text{and}\,\, \left(\frac{q}{p-\alpha} \times \frac{q}{p-\beta}\right)$

So, the sum of roots$\,=\,\left(\frac{q}{p-\alpha}+\frac{q}{p-\beta}\right)\\=\frac{q(p-\beta)+q(p-\alpha)}{(p-\alpha)(p-\beta)}\\=\frac{pq-q\beta+pq-q\alpha}{p^2-(\alpha+\beta)p+\alpha\beta}\\=\frac{2pq-q(\alpha+\beta)}{p^2-p^2+q}\\=\frac{2pq-pq}{q}\,\,\,[\text{Since,}\,\,\alpha+\beta=p]\\=p \cdots(3)$

The product of roots $\,=\left(\frac{q}{p-\alpha} \times \frac{q}{p-\beta}\right)\\=\frac{q^2}{p^2-(\alpha+\beta)p+\alpha\beta}\\=\frac{q^2}{p^2-p^2+q}\\=\frac{q^2}{q}\\=q\cdots (4)$

Hence by (2),(3) and (4), we get the required equation, $\,\,x^2-px+q=0.$

Reason for two equations being identical: 

Since $\,\,\alpha \,\,$  is a root of (1), we get $\,\,\alpha^2-p\alpha+q=0 \\ \Rightarrow  \alpha (p-\alpha)=q \\ \Rightarrow \alpha=\frac{q}{p-\alpha}$

Similarly, $\,\,\beta=\frac{q}{p-\beta}.$

So, the  roots of equation (1) are $\,\,\alpha,\,\,\beta\,\,$ which are equal or identical with $\,\,\frac{q}{p-\alpha} \,\,\text{and}\,\, \frac{q}{p-\beta}\,\,$ respectively and hence ​the result.

5. If $\,\,\frac{p^2}{q}\,\,\text{and}\,\,\frac{q^2}{p}\,\,$ are the roots of the equation $\,2x^2+7x-4=0,\,\,$ find the equation whose roots are $\,p\,$ and $\,q\,\,(p+q \,\,\text{real.})$

Sol. If $\,\,\frac{p^2}{q}\,\,\text{and}\,\,\frac{q^2}{p}\,\,$ are the roots of the equation $\,2x^2+7x-4=0, \\ \therefore \frac{p^2}{q} +\frac{q^2}{p}= -\frac 72 ,\\ \text{and}\,\,\,\frac{p^2}{q}  \times \frac{q^2}{p}= -\frac 42,\\ \Rightarrow pq=-2 \cdots (1)$

Now, $\frac{p^2}{q} +\frac{q^2}{p}=-\frac 72 \\ \Rightarrow p^3+q^3=7 \,\,[\text{Since,}\,pq=-2]\\ \Rightarrow (p+q)^3-3pq(p+q)=7 \\ \Rightarrow (p+q)^3+6(p+q)-7=0\\ \Rightarrow m^3+6m-7=0\,\,[m=p+q]\\ \Rightarrow m^3-m^2+m^2-m+7m-7=0 \\ \Rightarrow m^2(m-1)+m(m-1)+7(m-1)=0 \\ \Rightarrow (m-1)(m^2+m+7)=0 \\ \Rightarrow m-1=0,\,\,m^2+m+7=0 \\ \Rightarrow m=1,\,\,m=\frac{-1\pm \sqrt{1-4 \times 1 \times 7}}{2} \not \in \mathbb R$ 

Next, $\,\,m=1\,\,\,$ gives $\,\,p+q=1 \cdots(2)$

Hence, from (1) and (2), we get the required equation (with roots $\,\,p,\,\,q)\,:x^2-(p+q)x+pq=0 \\ \Rightarrow x^2-x-2=0 \\ [\text{Since,}\,\,p+q=1,\,\,pq=-2.]$ 

6. If $\,\alpha,\,\,\beta\,\,$ be the roots of the equation $\,x^2+2px-2q^2=0\,\text{where}\,\,p,q\,$ are rational and $\,\,p^2+q^2\,\,$ is not a perfect square, form the quadratic equation whose one root is $\,\,\alpha+\beta+\sqrt{\alpha^2+\beta^2}.$

Sol. If $\,\alpha,\,\,\beta\,\,$ be the roots of the equation $\,x^2+2px-2q^2=0,\cdots(1)\\ \therefore \alpha+\beta=-2p \\ \text{and}\,\,\alpha\beta=-2q^2.\\ \therefore \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\\~~~~~~~~~~~~~~~~~=4(p^2+q^2) \\ \therefore \sqrt{\alpha^2+\beta^2}=\pm2\sqrt{p^2+q^2}$

Hence, the roots of (1) is : $\,(-2p \pm2\sqrt{p^2+q^2})$

So, the required quadratic equation is given by : $\,\,x^2-(-2p +2\sqrt{p^2+q^2}\\-2p -2\sqrt{p^2+q^2})x+(-2p +2\sqrt{p^2+q^2}) \\ \times  (-2p -2\sqrt{p^2+q^2})=0 \\ \Rightarrow x^2+4px+4p^2-4p^2-4q^2=0 \\ \Rightarrow x^2+4px-4q^2=0.$

7. If $\,\,\alpha,\, \beta\,$ be the roots of the equation $\,cx^2+2bx+2c=0\,\,$ where $\,b,c\,$ are real and $\,\,c^2>b^2,\,$ find the quadratic equation whose one root is $\,\alpha+\beta+\sqrt{\alpha^2+\beta^2}.$

Sol. Since $\,\,\alpha,\, \beta\,$ be the roots of the equation $\,cx^2+2bx+2c=0 \\ \therefore \alpha+\beta=-\frac{2b}{c},\\ \text{and}\,\,\alpha \beta=\frac{2c}{c}=2$

Now, $\,\,\alpha^2+\beta^2\\=(\alpha+\beta)^2-2\alpha\beta\\=\frac{4b^2}{c^2}-4 \\=\frac{4(b^2-c^2)}{c^2} \\ \therefore \sqrt{\alpha^2+\beta^2}=\pm i \frac{2}{c}\sqrt{c^2-b^2}\,\,[\text{Since},\,c^2>b^2]$ 

Next, we have to find the quadratic equation whose one root is $\,\alpha+\beta+\sqrt{\alpha^2+\beta^2}.$

So, the roots of the quadratic equation is given by : $\,\,\alpha+\beta+\sqrt{\alpha^2+\beta^2}\\=-\frac{2b}{c} \pm i \frac 2c\sqrt{c^2-b^2}$

Hence, the required quadratic equation is given by: 

$x^2-\left[(-\frac{2b}{c}+i \frac 2c\sqrt{c^2-b^2})\\+(-\frac{2b}{c}-i \frac 2c\sqrt{c^2-b^2})\right]x\\+(-\frac{2b}{c}+i \frac 2c\sqrt{c^2-b^2})\\ \times (-\frac{2b}{c}-i \frac 2c\sqrt{c^2-b^2})=0 \\ \Rightarrow x^2+\frac{4bx}{c}+\frac{4b^2}{c^2}+\frac{4}{c^2}(c^2-b^2)=0 \\ \Rightarrow x^2+\frac{4bx}{c}+4=0 \\ \Rightarrow cx^2+4bx+4c=0.$