QUADRATIC EQUATIONS (Part-1) | S.N. Dey Math Solution Series

QUADRATIC EQUATIONS (Part-1) | S.N. Dey Math Solution Series



 Today  we are going to start new chapter : Quadratic Equation as a part of our S.N. Dey Math solution series.    

An equation of one unknown quantity (say $\,x\,$) in the form $\,ax^2+bx+c = 0\,$ is called a quadratic equation or an equation of the second degree. Here $\,a, b, c\,$ are constants and $\, a \neq 0\,\,$ while $\,b\,$ and $\,c\,$ may be zero; $\,a, b\,$ and $\,c\,$ are called the coefficient of $\,x^2\,$, the coefficient of $\,x\,$ and the constant (or absolute) term respectively. The values of $\,x\,$ which satisfy the equation are called the roots of the quadratic equation.

In this chapter, our endeavour will be to find the number of roots of a quadratic equation, nature of these roots and the manner in which they are related with the coefficients $\,a, b\,$ and the constant term $\,c\,$.


1.  If $\,\alpha\,$ and $\,\beta\,$ are the roots of the equation $\,x(2x-1)=1\,$, find the value of $\,\alpha^2-\beta^2\,\,$ and form the equation whose roots are $\,2\alpha-1\,$  and $\,2\beta-1.\,$

Sol. We have, the equation $\,x(2x-1)=1 \Rightarrow 2x^2-x-1=0 \cdots(1). \,$

Since $\,\alpha\,$ and $\,\beta\,$ are the roots of the equation (1), $\alpha+ \beta =\frac 12 ,\,\, \alpha. \beta=-\frac 12.$

Now, $\,\,(\alpha-\beta)^2=(\alpha+\beta)^2 -4\alpha \beta\\ \Rightarrow (\alpha- \beta)^2=(\frac 12)^2-4(-\frac 12)\\ \Rightarrow (\alpha-\beta)^2=\frac 14+2 =\frac 94 \\ \Rightarrow (\alpha-\beta)= \pm \frac 32. \\ \therefore \alpha^2-\beta^2\\=(\alpha+\beta)(\alpha-\beta)\\=\frac 12 \times (\pm \frac 32)\\=\pm \frac 34.$

To find the equation whose roots are $\,2\alpha-1\,$  and $\,2\beta-1 ,\,$ we first calculate the value of $\,\,(2 \alpha-1)(2 \beta-1)\,$ and $\,(2 \alpha-1)+(2 \beta-1).$

Now, $\,\,(2 \alpha-1)(2 \beta-1)\\=4 \alpha\beta-2(\alpha+\beta)+1 \\=4 \times (- \frac 12)-2 \times \frac 12+1\\=-2-1+1\\=-2 \cdots(2)$

Also,  $\,(2 \alpha-1)+(2 \beta-1)\\=2(\alpha+\beta)-2\\=2 \times (\frac 12) -2\\=-1 \cdots(3)$

Hence , the required equation is given by $\,\,x^2-[(2 \alpha-1)+(2 \beta-1)]x\\+(2 \alpha-1)(2 \beta-1)=0 \\ \Rightarrow x^2-(-1)x-2=0 \,\,[\text{By (2) and (3)}] \\ \Rightarrow x^2+x-2=0.$

2. If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,5x^2 + 7x+3=0\,$, find the value of $\,\,\frac{\alpha^3+\beta^3}{\alpha^{-1}+\beta^{-1}}.$

Sol. If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,5x^2 + 7x+3=0\,$,

then $\,\,\alpha+\beta=-\frac 75,\,\,\alpha \beta=\frac 35.$

Now, $\,\alpha^3+\beta^3\\=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)\\=(-\frac 75)^3-3 \times \frac 35 \times (-\frac 75)\\=-\frac{343}{125}+\frac{63}{25}\cdots(1)$

Next, $\,\,\alpha^{-1}+\beta^{-1}\\=\frac {1}{\alpha}+\frac{1}{\beta}\\=\frac{\beta+\alpha}{\alpha \beta}\\=\frac{-7/5}{3/5}\\=-\frac{7}{3} \cdots(2)$

Hence, $\,\,\frac{\alpha^3+\beta^3}{\alpha^{-1}+\beta^{-1}}\\=\frac{-\frac{343}{125}+\frac{63}{25}}{-7/3} \,\,[\text{By (2) and (3)}]\\=\frac{-28/125}{-7/3} \\=\frac{28}{125} \times \frac{3}{7}\\=\frac{12}{125}.$

3. If $\,p\,$ and $\,q\,$ are the roots of the equation $\,ax^2+bx+c=0\,$, find the value of $\,\,\frac{1}{(ap^2+c)^2}+\frac{1}{(aq^2+c)^2}.$

Sol. If $\,p\,$ and $\,q\,$ are the roots of the equation $\,ax^2+bx+c=0 \\  \therefore p+q=-\frac ba \cdots(1) \\pq=\frac ca \cdots(2)$

Also, since $\,p\,$ and $\,q\,$ are the roots of the equation $\,ax^2+bx+c=0 \\ \Rightarrow ap^2+bp+c=0 \\ \Rightarrow ap^2+c=-bp \cdots(3)\\ \text{Similarly,}\,\,aq^2+c=-bq \cdots(4) $

Now.$\,\,\frac{1}{(ap^2+c)^2}+\frac{1}{(aq^2+c)^2} \\=\frac{1}{b^2p^2}+\frac{1}{b^2q^2}\,\,[\text{By (3) and (4)}]\\= \frac{q^2+p^2}{(bpq)^2}\\=\frac{(p+q)^2-2pq}{b^2(pq)^2}\\=\frac{b^2/a^2-(2c/a)}{b^2. \frac{c^2}{a^2}}\,\,[\text{By (1) and (2)}]\\=\frac{b^2-2ca}{b^2c^2}$

4. If $\,\alpha,\beta\,$ and $\,\gamma,\delta\,$ are the roots of the equations $\,x^2-bx+c=0\,$ and $\,x^2- px +q=0\,$ respectively, show that, $\,(\alpha-\gamma)(\beta-\delta)+(\beta-\gamma)(\alpha-\delta) \\=2(c+q)-bp$.

Sol. Since $\,\alpha,\beta\,$ are roots of the equations $\,x^2-bx+c=0, \\ \therefore \alpha+\beta=b,\,\,  \alpha \beta=c.$  

Again, since $\,\gamma,\delta\,$ are roots of the equations $\,x^2- px +q=0, \\ \therefore \gamma+\delta=p,\,\,  \gamma \delta=q.$ 

Now, $\,(\alpha-\gamma)(\beta-\delta)+(\beta-\gamma)(\alpha-\delta) \\=\alpha \beta-\alpha \delta-\beta \gamma+\gamma \delta +\alpha \beta -\beta \delta-\gamma \alpha+\gamma \delta \\=2 \alpha \beta+2 \gamma \delta -\delta(\alpha+\beta)-\gamma(\alpha+\beta)\\=2(\alpha \beta+\gamma \delta)-(\alpha+\beta)(\gamma+\delta)\\=2(c+q)-bp$

Hence, the result follows.

5. If the roots of the equation $\,ax^2 - bx+ a=0\,$ be $\,\alpha, \, \beta,\,\,$ show that the equation with roots $\,\alpha^2+1 \,$ and $\,\beta^2+1\,$ will be  $\,a ^2x^2-b^2x+ b^2 = 0.$ 

Sol. The roots of the equation $\,ax^2 - bx+ a=0\,$ be $\,\alpha, \, \beta .$

Hence, $\,\,\alpha+\beta=\frac ba,\,\,\alpha \beta= \frac aa=1.$

Now, we calculate the values of  $\,(\alpha^2+1) +(\beta^2+1)$ and  $\,(\alpha^2+1) \times (\beta^2+1).$

$\therefore(\alpha^2+1) +(\beta^2+1)\\=\alpha^2+\beta^2+2\\=(\alpha+\beta)^2-2 \alpha\beta+2\\=\frac{b^2}{a^2}-2+2\\=\frac{b^2}{a^2} \\ \text{and,}\,\,(\alpha^2+1)(\beta^2+1)\\=\alpha^2\beta^2+\alpha^2+\beta^2+1\\=1^2+ (\alpha+\beta)^2-2 \alpha\beta+1\\=1+\frac{b^2}{a^2}-2+1\\=\frac{b^2}{a^2} $

Hence, the required quadratic equation is given by : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0\\ \Rightarrow x^2-\frac{b^2}{a^2}x+\frac{b^2}{a^2}=0 \\ \Rightarrow a^2x^2-b^2x+b^2=0.$


6. If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,2x^2+x+1=0,\,$ find the equation whose roots are $\,\frac{\alpha^2}{\beta}$ and $\,\frac{\beta^2}{\alpha}.$

Sol.  $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,2x^2+x+1=0.$

Hence, $\,\, \alpha+\beta =-\frac 12,\\ \text{and}\,\, \alpha\beta=\frac 12 \cdots(1)$

Now we have to find the equation whose roots are $\,\frac{\alpha^2}{\beta}$ and $\,\frac{\beta^2}{\alpha}.$

For that let us first compute $\,(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}) ,\,\,\,(\frac{\alpha^2}{\beta})(\frac{\beta^2}{\alpha})=\alpha\beta.$

Now, $\,\,\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\\=\frac{\alpha^3+\beta^3}{\alpha\beta}\\=\frac{(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}{1/2}\\=\frac{(-1/2)^3-3(1/2)(-1/2)}{1/2}\\=\frac{-\frac 18+\frac 32 \times \frac 12}{\frac 12}\\=\frac{(-\frac 18+\frac 32 \times \frac 12)\times 8}{\frac 12 \times 8}\\=\frac{-1+6}{4}\\=\frac 54 \cdots(2)$

Hence,  the required equation whose roots are $\,\frac{\alpha^2}{\beta}$ and $\,\frac{\beta^2}{\alpha},\,\,$ is given by : $x^2-(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}) x+(\frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha})=0 \\ \Rightarrow x^2-\frac 54 x+ \frac 12=0 \,\,[\text{By (1) and (2)}]\\ \Rightarrow 4x^2-5x+2=0.$ 

7. If $\,\alpha\,$ and $\,\beta\,$ are the roots of the equation $\,x^2+\alpha x+\beta=0,\,\,$ then find the numerical values of $\,\alpha\,$ and $\,\beta\, [\,\,\text{Here,}\,\, \alpha \neq \beta,\,\alpha \neq 0, \beta \neq 0.]$ 

Sol.  If $\,\alpha\,$ and $\,\beta\,$ are the roots of the equation $\,x^2+\alpha x+\beta=0,\,\,$ then $\,\alpha+\beta =-\alpha \cdots (1) \\ \text{and}\,\, \alpha \beta=\beta \\ \Rightarrow \alpha=1\,\,[\text{Since,}\,\beta \neq 0]$

Now, putting the value of $\,\alpha=1,\,$ in (1), we get $\,1+\beta=-1\\ \Rightarrow \beta=-2. $

8. If the roots of the equation $\,ax^2 + bx+c = 0\,\,$ be $\,\alpha\,$ and $\,\beta\,$, find the equation whose roots are $\,\alpha+\frac{\alpha^2}{\beta}\,$ and $\,\beta+\frac{\beta^2}{\alpha}$.

Sol. If the roots of the equation $\,ax^2 + bx+c = 0\,\,$ be $\,\alpha\,$ and $\,\beta\,$, then $\,\,\alpha+\beta=-\frac ba, \\  \text{and}\,\, \alpha \beta=\frac ca.$

Now to find the equation whose roots are $\,\alpha+\frac{\alpha^2}{\beta}\,$ and $\,\beta+\frac{\beta^2}{\alpha}$, we have to calculate $\,\,(\alpha+\frac{\alpha^2}{\beta})+(\beta+\frac{\beta^2}{\alpha}),\,\,\text{and}\,\, (\alpha+\frac{\alpha^2}{\beta})(\beta+\frac{\beta^2}{\alpha}).$

Now, the sum of roots : $\,\,(\alpha+\frac{\alpha^2}{\beta})+(\beta+\frac{\beta^2}{\alpha})\\=(\alpha+\beta)+(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha})\\=-\frac ba+(\frac{\alpha^3+\beta^3}{\alpha\beta})\\=-\frac ba+\frac{(\alpha+\beta)^3-3 \alpha\beta(\alpha+\beta)}{\alpha \beta}\\=-\frac ba+\frac{\left(-\frac ba\right)^3-3. \frac ca. \left(-\frac ba\right)}{\frac ca}\\=-\frac ba+\frac{-b^3+3abc}{a^2c}\\=\frac{-abc-b^3+3abc}{a^2c}\\=-\frac{b(b^2-2ac)}{a^2c}$

Product of the roots : $ \,(\alpha+\frac{\alpha^2}{\beta})(\beta+\frac{\beta^2}{\alpha})\\=\alpha\beta+\beta^2+\alpha^2+\alpha\beta \\=\alpha^2+2\alpha\beta+\beta^2\\=(\alpha+\beta)^2\\=\frac{b^2}{a^2}$

Hence,  the required equation whose roots are $\,\alpha+\frac{\alpha^2}{\beta}\,$ and $\,\beta+\frac{\beta^2}{\alpha}\,\,$ is : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0\\ \Rightarrow x^2+\frac{b(b^2-2ac)}{a^2c}x+\frac{b^2}{a^2}=0 \\ \Rightarrow a^2cx^2+b(b^2-2ac)x+b^2c=0 $

9. If $\,\,\alpha,\beta\,\,$ be the roots of the equation $\,\,x^2+px+q=0,\,$ show that $\,\,\frac{1}{\alpha+\beta}\,\,$ and $\,\,\frac{1}{\alpha}+\frac{1}{\beta}\,$ are the roots of $\,pqx^2+(p^2+q)x+p=0.$

Sol.  If $\,\,\alpha,\beta\,\,$ be the roots of the equation $\,\,x^2+px+q=0,\\ \alpha+\beta=-p,\\ \alpha\beta=q.$

Next, to find the equation whose roots are $\,\, \alpha_1=\frac{1}{\alpha+\beta}\,\,$ and $\,\,\beta_1=\frac{1}{\alpha}+\frac{1}{\beta}\,$ , we have to calculate $\,(\alpha_1+\beta_1),\,\,(\alpha_1.\beta_1).$

Now, $\,\alpha_1+\beta_1=\frac{1}{\alpha+\beta}+(\frac{1}{\alpha}+\frac{1}{\beta})\\~~~~~~~~~~~~~= \frac{1}{\alpha+\beta}+\frac{\alpha+\beta}{\alpha \beta}\\~~~~~~~~~~~~~=-\frac 1p-\frac pq\\~~~~~~~~~~~~~=-\frac{p^2+q}{pq} \\ \alpha_1.\beta_1=\frac{1}{\alpha+\beta}\times (\frac{1}{\alpha}+\frac{1}{\beta})\\~~~~~~~~~~=\frac{1}{\alpha+\beta} \times \frac{\alpha+\beta}{\alpha\beta}\\~~~~~~~~~~=\frac{1}{\alpha\beta}\\ ~~~~~~~~~~=\frac 1q$

Hence,  the required equation : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0\\ \Rightarrow x^2+(\frac{p^2+q}{pq})x+\frac 1q=0 \\ \Rightarrow pqx^2+(p^2+q)x+p=0$

10. If the roots of the equation $\,\,ax^2+bx+c=0\,\,$ be $\,\alpha\,$ and $\,\beta\,$, find the equation whose roots are $\,\,\frac{1}{\alpha}+1\,$ and  $\,\,\frac{1}{\beta}+1\,$.

Sol.  If $\,\,\alpha,\beta\,\,$ be the roots of the equation $\,\,ax^2+bx+c=0,\\ \alpha+\beta=-\frac ba,\\ \alpha\beta=\frac ca.$

Next, to find the equation whose roots are $\,\, \alpha_1=\frac{1}{\alpha}+1\,\,$ and $\,\,\beta_1=\frac{1}{\beta}+1\,$ , we have to calculate $\,(\alpha_1+\beta_1),\,\,(\alpha_1.\beta_1).$

Now, $\,\alpha_1+\beta_1\\=\frac{1}{\alpha}+1+\frac{1}{\beta}+1\\=\frac{1}{\alpha}+\frac{1}{\beta}+2\\=\frac{\alpha+\beta}{\alpha\beta}+2\\=\frac{-\frac ba}{\frac ca}+2\\=-\frac{b}{c}+2\\=-\frac{b-2c}{c} \\ \alpha_1.\beta_1\\=(\frac{1}{\alpha}+1)\times (\frac{1}{\beta}+1)\\=\frac{1}{\alpha \beta}+ \frac{1}{\alpha}+\frac{1}{\beta}+1\\=\frac ac+\frac{\alpha+\beta}{\alpha\beta}+1\\=\frac ac+\frac{-\frac ba}{\frac ca}+1\\=\frac ac-\frac bc+1\\=\frac{a-b+c}{c}$

Hence,  the required equation : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2+\left(\frac{b-2c}{c}\right)x+\left(\frac{a-b+c}{c}\right)=0 \\ \Rightarrow cx^2+(b-2c)x+(a-b+c)=0$

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