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QUADRATIC EQUATIONS (Part-2) | S.N. Dey Math Solution Series

 

QUADRATIC EQUATIONS (Part-2) | S.N. Dey Math Solution Series



In the previous article, we have discussed few Short Answer Type Questions (From Qst. 1-10) of S.N. Dey Math Exercise. In this article, we will discuss some more  problems and their solutions. So, let's start.

11. Form the quadratic equation whose roots $\,\alpha\,$ and $\,\beta\,$ satisfy the relations $\,\, \alpha \beta=768\,\,$ and $\,\,\alpha^2+\beta^2=1600.$ 

Sol.  $\,\,\alpha^2+\beta^2=1600 \\ \Rightarrow (\alpha+\beta)^2-2\alpha \beta=1600 \\ \Rightarrow (\alpha+\beta)^2=1600+ 2\times 768\,\,[\text{where,}\,\,\alpha\beta=768] \\ \Rightarrow \alpha+\beta=\pm \sqrt{3136}=\pm 56$ 

Hence, the quadratic equation whose roots $\,\alpha\,$ and $\,\beta\,$ satisfying the given relations is :  $\,x^2-(\alpha+\beta)x+\alpha\beta=0 \\ \Rightarrow x^2-(\pm 56)x+768=0 \\ \Rightarrow x^2 \mp 56x+768=0.$  

12. Form the quadratic equation whose roots are the squares of the roots of $\,\,x^2+3x+2=0.$

Sol. We have, $\,\,x^2+3x+2=0 \cdots (1) \\ \Rightarrow (x+1)(x+2)=0 \\ \Rightarrow x=-1,-2  $

The roots of (1), is $\,-1,-2.\,$

Now, we have to form an equation whose roots are $\,(-1)^2=1,\,(-2)^2=4.\cdots(2)$

Hence,  the required equation : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-(1+4)x+1\times 4=0 \\ \Rightarrow x^2-5x+4=0.$

13. Form an equation whose roots are the reciprocals of the roots of  $\,x^2+3x+4=0.$

Sol. Let the roots of $\,x^2+3x+4=0,\,\,$ is $\,\,\alpha\,$ and $\,\,\beta.$

So, $\,\,\alpha+\beta=-3,\,\, \alpha\beta=4 \cdots (1)$ Now, we have to form an equation whose roots are $\,\frac{1}{\alpha},\,\,\frac{1}{\beta}.$ 

Now, the sum of the roots $=\frac{1}{\alpha}+\frac{1}{\beta}\\=\frac{\beta+\alpha}{\alpha\beta}\\=-\frac{3}{4}\,\,[\text{By (1)}]\cdots(2)$

The product of the roots $=\frac{1}{\alpha} \times \frac{1}{\beta}\\=\frac{1}{4} \cdots(3)$

Hence,  the required equation : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-(-3/4)x+1/4=0 \\ \Rightarrow 4x^2+3x+1=0.$

14. If $\,3a^2=4a-5,\,\,3b^2=4b-5,\,\,[\text{where}\,\, a \neq b],\,\,$ find the value of $\,a^2+b^2.$

Sol. We have , $\,\,3a^2=4a-5 \cdots(1)\\3b^2=4b-5,\,\,[\text{where}\,\, a \neq b]\cdots(2)$

From (1) and (2), we can say $\,\,a,b\,\,$ are the roots of $\,3x^2-4x+5=0.\,\,$ 

So, $\,a+b=\frac 43,\,\, ab=\frac 53.\\ \therefore a^2+b^2\\=(a+b)^2-2ab\\=(4/3)^2-2\times (5/3)\\=\frac{16}{9}-\frac{10}{3}\\=\frac{16-30}{9}\\=-\frac{14}{9}.$ 

15. If $\,3p^2=5p+2\,\,\,\text{and}\,\, 3q^2=5q+2\,\,\text{where}\,\,p \neq q,\,\,$ obtain the equation whose roots are $\,\,(3p-2q)\,\,\text{and}\,\,(3q-2p).$

Sol. We have , $\,3p^2=5p+2 \cdots(1)\\\text{and}\,\, 3q^2=5q+2\,\,[\text{where}\,\,p \neq q]\cdots(2)$

From (1) and (2), we can say $\,p,\,q\,$ are the roots of $\,3x^2-5x-2=0. \\ \therefore p+q=\frac 53,\,\,\,\, pq=-\frac 23.$

Now, we have to obtain the equation whose roots are $\,\,(3p-2q)\,\,\text{and}\,\,(3q-2p).$

So, the sum of the roots $=(3p-2q)+(3q-2p)\\=3(p+q)-2(p+q)\\=p+q\\=\frac 53. $

The product of roots $=(3p-2q)(3q-2p)\\=13pq-6(p^2+q^2)\\=13\times (-\frac 23)-6[(p+q)^2-2pq]\\=-\frac{26}{3}-6[(5/3)^2-2 \times (-\frac 23)]\\=-\frac{26}{3}-6[\frac{25}{9}+\frac 43]\\=-\frac{26}{3} -6 \times \frac{37}{9}\\=-\frac{100}{3}$

Hence,  the required equation is : 

$\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-(5/3)x-\frac{100}{3}=0 \\ \Rightarrow 3x^2-5x-100=0.$

16. If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,x^2-4x+10=0,\,\,$ find the equation whose roots are $\,\,\frac{\alpha}{1+\beta}\,$ and $\,\,\frac{\beta}{1+\alpha}.$

Sol. Since $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,x^2-4x+10=0,\\ \alpha+\beta=-(-4)=4 \\ \alpha \beta=10.$

Now, we have to find the equation whose roots are $\,\,\frac{\alpha}{1+\beta}\,$ and $\,\,\frac{\beta}{1+\alpha}.$

So, the sum of the roots $=\,\frac{\alpha}{1+\beta}+\frac{\beta}{1+\alpha}\\=\frac{\alpha(1+\alpha)+\beta(1+\beta)}{(1+\alpha)(1+\beta)}\\=\frac{\alpha+\alpha^2+\beta+\beta^2}{(1+\alpha)(1+\beta)}\\=\frac{(\alpha+\beta)+(\alpha^2+\beta^2)}{(1+\alpha)(1+\beta)}\\=\frac{(\alpha+\beta)+(\alpha+\beta)^2-2\alpha\beta}{(1+\alpha+\beta+\alpha\beta)}\\=\frac{4+16-20}{1+4+10}\\=0.$

Product of the roots $=\frac{\alpha}{1+\beta} \times \frac{\beta}{1+\alpha}\\=\frac{\alpha \beta}{1+\alpha+\beta+\alpha\beta}\\=\frac{10}{1+4+10}\\=\frac{2}{3}$

Hence,  the required equation is : 

$\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-0.x+\frac 23=0 \\ \Rightarrow 3x^2+2=0.$

17. If $\,p\,$ and $\,q\,$ are the roots of the equation $\,\,3x^2+6x+2=0,\,\,$ show that the equation whose roots are $\,\,\left(-\frac{p^2}{q}\right)\,$ and $\,\,\left(-\frac{q^2}{p}\right)\,$ is $\,3x^2-18x+2=0.$ 

Sol. Since $\,p\,$ and $\,q\,$ are the roots of the equation $\,\,3x^2+6x+2=0, \\ p+q=-\frac 63=-2,\\ pq=\frac 23.$

Next, to find the equation whose roots are $\,\,\left(-\frac{p^2}{q}\right)\,$ and $\,\,\left(-\frac{q^2}{p}\right)\,$ , we need to calculate $\,\, -\frac{p^2}{q}-\frac{q^2}{p},\,$ and $\,\left(-\frac{p^2}{q}\right) \times \left(-\frac{q^2}{p}\right)=-pq=-\frac 23 \cdots(1)$ 

Now, $-\frac{p^2}{q}-\frac{q^2}{p}\\=-(\frac{p^3+q^3}{pq})\\=-\frac{(p+q)^3-3pq(p+q)}{pq}\\=-\frac{(-2)^3-3 \times (2/3) \times (-2)}{2/3}\\=\frac{4 \times 3}{2}\\=6\cdots(2)$

Hence,  the required equation is : 

$\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-6x+\frac 23=0 \\ \Rightarrow 3x^2-18x+2=0.$

18. If the sum of the roots of a quadratic equation is $\,2\,$ and the sum of their cubes is $\,\,27,\,$ find the equation.

Sol. Let the roots of the quadratic equation be $\,\,\alpha,\,\beta.$ Then by question, we have $\,\, \alpha+\beta=2,\\\alpha^3+\beta^3=27. \\ \Rightarrow (\alpha+\beta)^3-3 \alpha\beta(\alpha+\beta)=27 \\ \Rightarrow 2^3-3 \alpha \beta \times 2=27 \\ \Rightarrow 8-6 \alpha\beta=27 \\ \Rightarrow -6\alpha\beta=27-8 \\ \Rightarrow \alpha\beta=-\frac{19}{6}$

Hence,  the required equation is : 

$\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-(\alpha+\beta)x+\alpha\beta=0 \\ \Rightarrow x^2- 2x-\frac{19}{6}=0 \\ \Rightarrow 6x^2-12x-19=0.$

19. If  the sum of roots of the equation $\,x^2-px+q=0\,$ be three times their difference, then show that $\,\,2p^2=9q.$

Sol.  Let $\,\alpha,\,\beta\,\,$ be the roots of the given equation, then by question , $\,\,\alpha+\beta=3(\alpha-\beta) \cdots(1)$

Next , by our supposition , we get $\,\alpha+\beta=-(-p)=p \cdots(2) \\ \alpha. \beta=q \cdots(3)$  

Now, by (1) and (2), we have , $\,\,3(\alpha-\beta)=p \\ \Rightarrow \alpha-\beta=\frac p3 \cdots(4)$

By $\,\,(2) \,\text{and}\, (4)$, we get $\,\,(\alpha+\beta)+(\alpha-\beta)=p +\frac p3 \\ \Rightarrow 2 \alpha=\frac{4p}{3}\\ \Rightarrow \alpha=\frac{2p}{3} \cdots(5) \\ \therefore (2p/3). \beta=q\,\,[\text{By (3)}]\\ \Rightarrow \beta=\frac{3q}{2p} \cdots(6)$

From $\,\,(4),\,(5),\,(6)\,$ we get , $\,\,\frac{2p}{3}-\frac{3q}{2p}=\frac p3 \\ \Rightarrow \frac{4p^2-9q}{6p}=\frac p3 \\ \Rightarrow 4p^2-9q=2p^2 \\ \Rightarrow 2p^2=9q.$

20. If the roots of the equation is $\,ax^2+bx+c=0\,$ are in the ratio $\,3:4\,$ show that $\,12b^2=49ac.$

Sol. By question , let $\,3\alpha,\,4\alpha\,\,$ be the roots of the equation is $\,ax^2+bx+c=0.$

Then , the sum of roots $=3 \alpha+4 \alpha=-\frac ba\\ \Rightarrow 7 \alpha=-\frac ba \\ \Rightarrow \alpha=-\frac{b}{7a} \cdots (1)$

The product of the roots $=(3 \alpha) \times (4 \alpha)=\frac ca\\ \Rightarrow 12 \alpha^2=\frac ca \\ \Rightarrow 12 (-\frac{b}{7a})^2=\frac ca \\ \Rightarrow 12 \times \frac{b^2}{49a^2}=\frac ca \\ \Rightarrow 12b^2=49ac.$

>>>>>>>>>>>>>>>>>>>>> To continue with the Quadratic Equations (Part-3), click here.

To get full PDF of S.N. Dey  Math Solutions on Quadratic Equations, click here

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