Differentiation (Part-38) | S N De

 

 In the previous article , we have discussed few Short Answer Type Questions (From Qst. 21-30) of S.N. Dey Math Exercise. In this article,    we are going to start new chapter : Quadratic Equations (Part-4) as a part of our S.N. Dey Math solution series.  So, let's start.

31. Show that the roots of the equation $\,(a^2-bc)x^2 + 2(b^2-ca)x+c^2-ab=0\,$ are equal if either $\,b=0\,\,$ or, $\,\,a^3 + b^3 + c^3-3abc = 0.$

Sol. Since the roots of the equation $\,(a^2-bc)x^2 + 2(b^2-ca)x+c^2-ab=0\,$ are equal $\,\,D =0 .$

Hence, $\,\,[2(b^2-ca)]^2-4(a^2-bc)(c^2-ab)=0 \\ \Rightarrow b^4-2ab^2c+c^2a^2-a^2c^2\\+a^3b+bc^3-ab^2c=0 \\ \Rightarrow b(a^3 + b^3 + c^3-3abc)=0 \cdots (1)$

So, from (1), we can conclude that either $\,b=0\,\,$ or, $\,\,a^3 + b^3 + c^3-3abc = 0.$

32. If one root of the equation $\,qx^2 + px+q=0 \,\,\,(p, q\,$ are real) be imaginary, show that the roots of the equation $\,\,x^2-4qx+p^2 = 0\,\,$ are real and unequal.

Sol. If one root of the equation $\,qx^2 + px+q=0 \cdots(1) \,\,\,(p, q\,$ are real) be imaginary, the other root must be conjugate complex numbers as we know , in a quadratic equation with real coefficients, imaginary roots occur in conjugate pairs. 

Since roots of (1) are imaginary, the discriminant of $\,\,(1) :D <0 \\ \therefore p^2-4q^2<0 \cdots (2)$

Now, the discriminant $\,(D')\,$ of the equation $\,\,x^2-4qx+p^2 = 0\,\,$ is : $\,(-4q)^2-4.1.p^2 \\=4(4q^2-p^2)>0\,\,\,[\text{By (2)}]$

Hence, he roots of the equation $\,\,x^2-4qx+p^2 = 0\,\,$ are real and unequal.

33. If the roots of the equation $\,qx^2 + 2px + 2q = 0\,\,$ are real and unequal , prove that, the roots of the equation $\,\,(p+q)x^2+2qx+(p-q) =0\,\,$ are imaginary.

Sol. Since the roots of the equation $\,qx^2 + 2px + 2q = 0\,\,$ are real and unequal , the discriminant $\,(D) >0 \\ \therefore (2p)^2-4.q.2q >0  \\ \Rightarrow p^2-2q^2 >0 \cdots(1)$

Again, the discriminant  $\,(D')\,$ of the equation $\,\,(p+q)x^2+2qx+(p-q) =0\,\,$ is : $\,(2q)^2-4.(p+q)(p-q) \\=4[q^2-p^2+q^2]\\=4(2q^2-p^2)<0\,\,[\text{By (1)}]$

Since  the discriminant of the equation $\,\,(p+q)x^2+2qx+(p-q) =0\,\,$ is $\,<0,\,\,$ the roots of the aforesaid equation are imaginary.

34. If the roots of the equation $\,\,x^2-2(a+b)x+a(a+2b+c)=0\,\,$ be equal, prove that, $\,a, b, c\,$ are in G.P.

Sol. Since the roots of the equation $\,\,x^2-2(a+b)x+a(a+2b+c)=0 \cdots (1)\,\,$ be equal, the discriminant of $\,\,(1)\, : D=0.$ 

Therefore, $\,\,[-2(a+b)]^2-4.1.a(a+2b+c) =0 \\ \Rightarrow 4(a^2+2ab+b^2-a^2-2ab-ac)=0 \\ \Rightarrow b^2=ac \cdots(1)$ 

From (1), the required result follows.

35. If the equations $\,\,x^2+px+q=0\,\,$ and $\,\,x^2+p'x+q'=0\,\,$ have a common root, prove that, it is either $\,\,\frac{pq'-p'q}{q-q'}\,\,\text{or,}\,\, \frac{q-q'}{p-p'}.$

Sol. If the equations $\,\,x^2+px+q=0\,\,$ and $\,\,x^2+p'x+q'=0\,\,$ have a common root, then let the common root be $\,\,\alpha. $

So, $\,\,\alpha^2+p\alpha+q=0 \\ \alpha^2+p'\alpha+q'=0. $

By cross-multiplication, we get $\, \frac{\alpha^2}{pq'-p'q}=\frac{\alpha}{q-q'}=\frac{1}{p'-p} \\ \Rightarrow \alpha^2=\frac{pq'-p'q}{p'-p} \cdots(1) ,\,\,\alpha=\frac{q-q'}{p'-p} \cdots(2)$

Next, from (1) and (2), we get $\,\, \alpha.\alpha=\frac{pq'-p'q}{p'-p} \\ \Rightarrow \alpha \times \frac{q-q'}{p'-p}=\frac{pq'-p'q}{p'-p} \\ \Rightarrow \alpha=\frac{pq'-p'q}{q-q'}$

Hence, the common root of the aforesaid equation is either $\,\,\frac{pq'-p'q}{q-q'}\,\,\text{or,}\,\, \frac{q-q'}{p-p'}.$

36. Prove that if the equations $\,\,x^2+px+r=0\,$ and $\,\,x^2+rx+p=0\,\,$ have a common root , then either $\,\,p=r\,$ or, $\,\,1+p+r=0.$

Sol. If the equations $\,\,x^2+px+q=0\,\,$ and $\,\,x^2+p'x+q'=0\,\,$ have a common root, then let the common root be $\,\,\alpha. $

So, $\,\,\alpha^2+p\alpha+q=0 \\ \alpha^2+p'\alpha+q'=0 $

Now, by cross-multiplication , we get $\, \frac{\alpha^2}{p^2-r^2}=\frac{\alpha}{r-p}=\frac{1}{r-p} \\ \Rightarrow \frac{\alpha^2}{-(p+r)}=\frac{\alpha}{1}=\frac 11 \,\,[ \text{Since,}\,\, r \neq p]\\ \Rightarrow \alpha^2=-(p+r) \,\text{and}\,\, \alpha=1. \\ \therefore 1^2=-(p+r) \\ \Rightarrow 1+p+r=0.$

37. If $\,x\,$ is real and the expression $\,\,3x^2-17x+20\,\,$ is always positive , show that, $\,x\,$ can not lie between $\,\,\frac 53\,\,$ and $\,\,4.$

Sol.   $\,\,3x^2-17x+20 >0\\ \Rightarrow 3x^2-12x-5x+20>0 \\ \Rightarrow 3x(x-4)-5(x-4)>0 \\ \Rightarrow (x-4)(3x-5)>0 $

Now, if $\,\,3x^2-17x+20 <0 \\ \Rightarrow \frac 53 <x<4 \cdots(1)$

From (1), we can say if $\,\,3x^2-17x+20 <0\,\,$ then $\,x\,$ must lie between $\,\,\frac 53 \,\text{and}\,\, 4.$

But since $\,\,3x^2-17x+20 >0,\,\,x $ can not lie between  between $\,\,\frac 53 \,\text{and}\,\, 4.$

38. Find the limits of real value of $\,x\,$ so that the expression $\,\,5x^2+6x-8\,\,$ is non-negative.

Sol. $\,\,5x^2+6x-8 \\=5x^2+10x-4x-8\\=5x(x+2)-4(x+2)\\=(x+2)(5x-4) \cdots(1)$

From (1), we can say either $\,\,(x+2) \leq 0,\,\,(5x-4) \leq 0 \cdots(2) \\ \text{or,}\,\,(x+2)\geq0,\,\,(5x-4) \geq 0 \cdots(3)$

From (2), we get $\,\, x \leq -2\,\,\,\text{and}\,\,x \leq \frac 45 \\ \Rightarrow -\infty <x \leq -2 \\ \Rightarrow x \in (-\infty,-2].$

From (3), we get $\,\, x \geq -2\,\,\text{and}\,\, x \geq \frac 45 \\ \Rightarrow \frac 45 \leq x < \infty \\ \Rightarrow x \in [\frac 45, \infty)$

Hence, the limits of real value of $\,x\,$ is : $\,\,(-\infty,-2] \cup [\frac 45, \infty).$

39. For any real positive value of $\,x\,$, show that $\,\,3-x \not > \frac{7}{x+2}.$

Sol. Let $\,\,y=3-x-\frac{7}{x+2} \\ \Rightarrow y=\frac{(3-x)(x+2)-7}{x+2}\\~~~~~~~=-\frac{x^2-x+1}{x+2}\\~~~~~~~=-\frac{x^2-2.x.\frac 12+(\frac 12)^2+\frac 34}{x+2}\\~~~~~~~=-\frac{-(x-\frac 12)^2+\frac 34}{x+2} \cdots(1)$

From (1), we can say that since $\,\,x>0,\,\,\text{and}\,\, \left(x-\frac 12\right)^2 \geq 0,\,\,$ so $\,\,y<0 \\ \Rightarrow y \not >0 \\ \Rightarrow 3-x \not > \frac{7}{x+2}.$

40. If $\,x\,$ is real can (i) the value of $\,\,3+2x-x^2\,\,$ be greater than $\,4\,\,\,$ and (ii) that of $\,\,x^2+3x+1\,\,$ less than $\,\,\left(-\frac 54\right)\,\,?$

(i) Sol. Let $\,\,y=3+2x-x^2 \\ \Rightarrow  x^2-2x-(3-y)=0 \cdots(1)$

Since, $\,x\,$ is real, the discriminant of (1) , $\,\,D \geq 0 \\ \Rightarrow (-2)^2-4.1.\{-(3-y)\} \geq 0 \\ \Rightarrow 4+4(3-y) \geq 0 \\ \Rightarrow 1+3-y \geq 0 \\ \Rightarrow y \leq 4.$

Hence, the value of $\,\,3+2x-x^2\,\,$ can not be greater than $\,4 .$ 

(ii) Sol. Let $\,y=x^2+3x+1 \cdots(2)\\ \Rightarrow x^2+3x+(1-y) =0$

Since , $\,\,x\,$ is real, the discriminant of (2) is : $\,\,D' \geq 0 \\ \therefore 3^2-4.1.(1-y) \geq 0 \\ \Rightarrow 1-y \leq \frac 94 \\ \Rightarrow y \geq -\frac 54.$

Hence, the value of $\,\,x^2+3x+1\,\,$ can not be less than $\,-\frac 54.$ 

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