Differentiation (Part-38) | S N De

 In the previous article , we have discussed few Short Answer Type Questions (From Qst. 11-20) of S.N. Dey Math Exercise. In this article,    we are going to start new chapter : Quadratic Equations (Part-3) as a part of our S.N. Dey Math solution series.  So, let's start.

21. If one root of the equation $\,\,x^2+(5a+2)x+5a+2=0\,$ is five times the other root, then find the numerical value of $\,a.$

Sol.  Let  the roots of the equation $\,\,x^2+(5a+2)x+5a+2=0\,$ are $\, \alpha,\,\,5 \alpha.$

Then, the sum of roots $=\alpha+5 \alpha=-(5a+2)/1\\ \Rightarrow \alpha=-(5a+2)/6 \cdots(1)$

The product of roots $=\alpha. 5\alpha=(5a+2)/1 \\ \Rightarrow 5 \alpha^2=5a+2 \\ \Rightarrow 5[-(5a+2)/6 ]^2=(5a+2) \\ \Rightarrow \frac{5}{36} \times (5a+2)=1 \\ \Rightarrow 5(5a+2)=36 \\ \Rightarrow 25a+10=36 \\ \Rightarrow 25a=36-10 \\ \Rightarrow a=\frac{26}{25}.$

22. Form the quadratic equation whose roots are the cubes of the roots of $\,\,x^2-4x+3=0.$

Sol. We have, $\,\,x^2-4x+3=0 \\ \Rightarrow x^2-3x-x+3=0 \\ \Rightarrow x(x-3)-1(x-3)=0 \\ \Rightarrow (x-3)(x-1)=0 \\ \Rightarrow x=3,1.$

We need to find an equation whose roots are $\,\,1^3=1,\,3^3=27.$

Hence, the required quadratic equation is : $\,x^2-(\text{sum of roots})x+\text{product of roots}=0 \\ \Rightarrow x^2-(1+27)x+1 \times 27=0 \\ \Rightarrow x^2-28x+27=0$

23. The ratio of the roots of the equation $\,\,ax^2+bx+c=0\,\,$ is $\,\,r:1.$ Prove that, $\,b^2r=ac(r+1)^2\,\,$ and hence find the condition so that the roots may be equal to each other.

Sol. Let $\,\,\alpha,\,\,\beta\,$ be the roots of the equation $\,\,ax^2+bx+c=0\\ \text{so that}\,\,\, \alpha+\beta=-\frac ba \cdots(1) \\  \alpha\beta=\frac ca \cdots(2)$ 

By the question, $\, \alpha : \beta=r:1 \\ \Rightarrow \alpha=\beta r\cdots(3)$

Hence, from $\,(1), (2)\,\, \text{and}\,\, (3)\,$, we get $\,\beta r \times \beta=\frac ca \\ \Rightarrow \beta^2 \times r=\frac ca\cdots(4)\\ \text{and}\,\,\,\beta r+\beta=-\frac ba  \\ \Rightarrow \beta(r+1)=-\frac ba\\ \Rightarrow \beta=-\frac{b}{a(r+1)} \cdots(5)$

From (4) and (5), we get $\,\,\frac{b^2}{a^2(r+1)^2} \times r= \frac ca \\ \Rightarrow b^2r=ac(r+1)^2.$

24. If the roots of the equation $\,\,px^2+rx+r=0\,$ are in the ratio $\,\,a:b,\,$ prove that , $\,\,p(a+b)^2=rab.$

Sol.  Let the roots of the equation $\,\,px^2+rx+r=0\,$ be $\,\,a \alpha,\, b\alpha \\ \text{so that},\,\, a\alpha+b\alpha=-\frac rp \\ \Rightarrow \alpha=-\frac{r}{p(a+b)} \\ \text{and}\,\, a\alpha \times b\alpha=\frac rp \\ \Rightarrow ab \times \alpha^2=\frac rp \\ \Rightarrow ab \times  \frac{r^2}{p^2(a+b)^2}=\frac rp \\ \Rightarrow p(a+b)^2=rab.$

25. In a quadratic equation of the form $\,\,x^2+mx+n=0\,\,$ the constant term is misprinted $\,56\,$ for $\,54\,$ and the roots are, therefore, obtained as $\,7\,$ and $\,8.\,$ Find the roots of the equation correctly printed.

Sol. By the question, $\,\,x^2+mx+56=0 \\ \Rightarrow x^2+mx+ 7 \times 8=0 \cdots(1).$

Next, $\,7\, \text{and}\,\,8\,$ being the roots of (1), we have $\,\,7+8=-\frac m1 \\ \Rightarrow m=-15.$

Now, the correctly printed equation can be written by placing $\,54\,$ instead of $\,56\,$ as the constant term. Hence, the equation correctly printed can be written as $\,\,x^2-15x+54=0 \\ \Rightarrow (x-6)(x-9)=0 \\ \Rightarrow x=6,9.$

So, the roots of the equation correctly printed  are $\,\, 6 \,\text{and}\,\,9.$

26. For what value of $\,m\,$ the roots of the equation $\,\,(m+1)x^2+2(m+3)x+m+8=0\,$ are equal?

Sol. The roots of the equation $\,\,(m+1)x^2+2(m+3)x+m+8=0\,$ are equal if $\,\,[2(m+3)]^2-4\times (m+1) \times (m+8)=0 \\ \Rightarrow 4(m^2+2 \times m \times 3+3^2)-4 \times (m^2+8m+m+8)=0 \\ \Rightarrow m^2+6m+9-(m^2+9m+8)=0 \\ \Rightarrow -3m+1=0 \\ \Rightarrow m=\frac 13$

27. If one root of the equation $\,x^2 + bx+8 = 0\,$ be $\,4\,$ and the roots of the equation $\,x^2 + bx+c= 0\,$ are equal, find the value of $\,c$.

Sol. Since one root of the equation $\,x^2 + bx+8 = 0\,$ be $\,4 ,\,$ so $\,\,4^2+b \times 4+8=0 \\ \Rightarrow 4b=-16-8 \\ \Rightarrow b=-\frac{24}{4}=-6$

Again, since  the roots of the equation $\,x^2 + bx+c= 0 \,\,\text{or,}\,\, x^2-6x+c=0\,\,$ are equal, $\,\, (-6)^2-4 \times 1 \times c=0 \\ \Rightarrow 36=4c \\ \Rightarrow c=\frac{36}{4}=9.$

28. For what value of $\,m\,$ the roots of the equation $\,\,\frac{3}{x+3+m} +\frac{5}{x+5+m}=1\,\,$ are equal in magnitude and opposite in signs?

Sol.  $\,\,\frac{3}{x+3+m} +\frac{5}{x+5+m}=1\\ \Rightarrow \frac{3(x+5+m)+5(x+3+m)}{(x+3+m)(x+5+m)}=1 \\ \Rightarrow x^2+(5+m+3+m)x+(3+m)(5+m)\\=8x+(30+8m) \\ \Rightarrow x^2+2mx+(m^2-15)=0 \cdots(1)$

Since the roots of the equation (1) are equal in magnitude and opposite in signs, if one root of (1) is $\,\alpha,\,\,$ the other root must be $\,\,-\alpha. $

Hence,$\,\, \alpha+(-\alpha)=2m \\ \Rightarrow m=0.$ 

29. Find the condition so that the roots of the equation $\,\,\frac{a}{x-a}+\frac{b}{x-b}=5\,\,$ are equal in magnitude but opposite in signs.

Sol. $\,\,\frac{a}{x-a}+\frac{b}{x-b}=5 \\ \Rightarrow \frac{a(x-b)+b(x-a)}{(x-a)(x-b)}=5 \\ \Rightarrow (a+b)x-2ab=5[x^2-(a+b)x+ab] \\ \Rightarrow 5x^2-6(a+b)x+7ab=0 \cdots(1)$ 

Since the roots of the equation (1) are equal in magnitude and opposite in signs, if one root of (1) is $\,\alpha,\,\,$ the other root must be $\,\,-\alpha. \\ \therefore \alpha+(-\alpha)=\frac{6(a+b)}{5} \\ \Rightarrow a+b=0.$ 

30. Show that the roots of each of the following equations are rational ($\,a, b, c\,$ are rational):

$\,(i) (b+c)x^2 - (a + b + c)x+ a = 0 \\(ii) (a-b+c)x^2 +2cx + (b+c-a) = 0$

(i) Sol. $\,\,(b+c)x² - (a + b + c)x+ a = 0 \cdots(1)$

The discriminant of (1) is : $\,\,D=[-(a+b+c)]^2- 4(b+c).a \\ ~~~~= (a+b+c)^2-4a(b+c)\\~~~~=(b+c-a)^2 \geq 0.$

Since the discriminant of (1) is $\, \geq 0,\,$ and $\,\,D\,\,$ is a perfect square, so the roots of (1) are rational. 

(ii) Sol. $\, (a-b+c)x^2 +2cx + (b+c-a) = 0 \cdots(2)$

The discriminant of (2) is : $\, D=(2c)^2-4 (a-b+c)(b+c-a) \\~~~~=4[c^2-\{c+(a-b)\}\{c-(a-b)\}]\\~~~~=4[c^2-c^2+(a-b)^2]\\~~~~=4(a-b)^2 \geq 0$

Since the discriminant of (2) is $\, \geq 0,\,$ and $\,\,D\,\,$ is a perfect square, so the roots of (2) are rational. 

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