Differentiation (Part-38) | S N De

  In the previous article , we have discussed Long Answer Type Questions (From Qst. 16-20) of S.N. Dey Math Exercise. In this article,  we are going to start Long Answer Type Questions of Quadratic Equations (Part-9) as a part of our S.N. Dey Math solution series.  So, let's start.

 21. If the roots of the equation $\,\,a(b-c)x^2+b(c-a)x+c(a-b)=0\,\,$ be equal, show that $\,\,\frac 1a+\frac 1c=\frac 2b.$

Sol. Since the roots of the equation $\,\,a(b-c)x^2+b(c-a)x+c(a-b)=0\,\,$ be equal, $\,b^2(c-a)^2-4[a(b-c)][c(a-b)]=0\\ \Rightarrow b^2(c-a)^2-4ac(ab-b^2-ac+bc)=0 \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\text{Since,}\,\,D=0]\\ \Rightarrow (bc-ab)^2+4.bc.ab+4a^2c^2-4ac(ab+bc)=0 \\ \Rightarrow (bc+ab)^2+(2ac)^2-2.2ac(ab+bc)=0 \\ \Rightarrow (bc+ab-2ac)^2=0\\ \Rightarrow bc+ab-2ac=0 \\ \Rightarrow bc+ab=2ac \\ \Rightarrow \frac 1a+\frac 1c=\frac 2b\,\,\,\text{(proved)}$

22. Find the condition that the roots of the quadratic equation $\,x^2+px+q=0\,$ should be (1) both positive (1) both negative (i) one is positive and the other negative (iv) equal in magnitude and opposite in signs (v) reciprocal to one another.

Sol. Let $\,\,\alpha,\,\beta\,$ be the roots of the quadratic equation $\,x^2+px+q=0 \\ \text{so that}\,\,\alpha+\beta=-p \\ \text{and}\,\,\alpha \beta=q.$

Now , if $ (i)\,\,\alpha,\beta>0,\,\,p<0 \,\,\text{and}\,\,q>0 \\ (ii)\,\,\,\,\,\alpha,\beta<0,\,\,p>0 \,\,\text{and}\,\,q>0 \\ (iii)\,\,\,\alpha >0,\beta<0,\,\text{or}\,\,\alpha<0,\,\beta>0 \,\,\text{then}\\ p>0\,\text{or}\,\,p<0,\,\,\text{but}\,\,q <0.\\ (iii)\, \,\,\alpha =-\beta,\,\, \text{or}\,\,\beta=-\alpha,\,\,\text{then} \\ p=0,\,\,q<0 \\ (v)\,\,\,\alpha=\frac{1}{\beta}\,\,\text{or}\,\,\beta=\frac{1}{\alpha},\\ \text{then}\,\,p>0\,\text{or}\,\,p<0,\,\,\text{but}\,\,q=1. $

23. If the roots of the equation $\,x^2+x+a=0\,$ be real and unequal, then prove that the roots of the equation $\,2x^2-4(1 + a)x+ 2a^2+3=0\,\,$ are imaginary ($\,a\,$ is real).

Sol. Since  the roots of the equation $\,x^2+x+a=0 \cdots(1)\,$ be real and unequal, so $\,\,D=1^2-4a>0 \\ \Rightarrow a<\frac 14.$

Now, $\,2x^2-4(1 + a)x+ 2a^2+3=0 \cdots(2)\,\,$ . The discriminant of (2) is : $\,\,D'=[-4(1+a)]^2-4 \times 2 \times (2a^2+3)\\=8\left(2+4a+2a^2-2a^2-3\right)\\=8(4a-1)<0\,\,[\text{Since,}\,\,a <\frac 14]$ 

Since the discriminant of (2) is negative, so the roots of (2) are imaginary.

24. Prove that, if the roots of the equation $\,(a^2+ b^2)x^2 + 2(bc + ad)x+ (c^2 + d^2)=0\,$ be real, then they are equal.

Sol. The discriminant of the equation $\,(a^2+ b^2)x^2 + 2(bc + ad)x\\+ (c^2 + d^2)=0 \cdots (1)\,$ : 

$\,\,D=[2(bc+ad)]^2-4(a^2+b^2)(c^2+d^2)\\=4(b^2c^2+2abcd+a^2d^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2)\\=-4(a^2c^2-2abcd+b^2d^2)\\=-4(ac-bd)^2$

Now, since the roots of (1) are real, so $\,\,-4(ac-bd)^2\geq 0\,\,$ but $\,\,-4(ac-bd)^2\not > 0\,\,$ and so $\,\,D=0\,$ and hence roots of (1) are equal.

25. Show that the roots of the $\,(a^4 + b^4)x^2 + 4abcdx + (c^4 + d^4) = 0\,\,$ cannot be different, if real.

Sol. The discriminant of the equation $\,\,(a^4 + b^4)x² + 4abcdx + (c^4 + d^4) = 0\,\cdots (1)\,$ : $\,\,D= (4abcd)^2-4(a^4 + b^4)(c^4 + d^4)\\=4(4a^2b^2c^2d^2-a^4c^4-a^4d^4-b^4c^4-b^4d^4)\\=-4[(a^2c^2-b^2d^2)^2+(a^2d^2-b^2c^2)^2]$

Since the roots of (1) are real, so $\,\,D \geq 0\,\,$ but in this case, $\,D \not>0\,\,$ and so $\,\,D=0\,\,\,$ and hence the roots can not be different if real.

26. (i) For what values of $\,m\,$ the equations $\,3x^2 + 4mx + 2 = 0\,$ and $\,2x^2 + 3x - 2 = 0\,$ will have a common root?

(ii) Show that the equations $\,px^2 +qx+r=0\,$ and $\,qx^2+rx+p=0\,$ will have a common an root if $\,\,p+q+r=0\,\,\text{ or,}\,\, p=q=r.$ 

Sol. (i) Let $\,\alpha\,$ be the common root of  the equations $\,3x^2 + 4mx + 2 = 0\,\cdots(1)$ and $\,2x^2 + 3x - 2 = 0\,\cdots(2). $

Hence, $\,3 \alpha^2+4m\alpha+2=0 \cdots(3) \\ 2 \alpha^2+3 \alpha-2=0 \cdots(4)$

by cross-multiplication, we get 

$\,\,\frac{\alpha^2}{-8m-6}=\frac{\alpha}{4+6}=\frac{1}{9-8m} \\ \Rightarrow \alpha^2=\frac{-(8m+6)}{9-8m},\,\cdots(5) \\ \alpha=\frac{10}{9-8m} \cdots(6)$

Now, from (5) and (6), we get $\,\,\left(\frac{10}{9-8m}\right)^2=\left(\frac{-(8m+6)}{9-8m}\right) \\ \Rightarrow 100=-(9-8m)(8m+6) \\ \Rightarrow 50=(8m-9)(4m+3) \\ \Rightarrow 32m^2-12m-77=0 \\ \Rightarrow (4m-7)(8m+11)=0 \\ \therefore m=\frac 74,\,\,-\frac{11}{8}.$

Sol. (ii) Let $\,\alpha\,$ be the common root of  the equations $\,\,px^2 +qx+r=0\,\,\cdots(1)$ and $\,qx^2+rx+p=0\,\cdots(2). $

So, $\,\,p\alpha^2+q\alpha+r=0 \cdots(3) \\ q\alpha^2+r \alpha+p=0 \cdots(4)$

From (3) and (4) we get by cross-multiplication, $\,\,\frac{\alpha^2}{pq-r^2}=\frac{\alpha}{qr-p^2}=\frac{1}{pr-q^2} \\ \Rightarrow \alpha^2=\frac{pq-r^2}{pr-q^2},\,\,\alpha=\frac{qr-p^2}{pr-q^2} \cdots (5)$

Hence, by (5) we get, $\,\left(\frac{qr-p^2}{pr-q^2} \right)^2=\frac{pq-r^2}{pr-q^2}\\ \Rightarrow (qr-p^2)^2=(pq-r^2)(pr-q^2)\\ \Rightarrow q^2r^2-2p^2qr+p^4=p^2qr-pq^3-pr^3+q^2r^2\\ \Rightarrow p^3+q^3+r^3-3pqr=0 \\ \Rightarrow (p+q+r)(p^2+q^2+r^2-pq-qr-rp)=0 \\ \text{so, either}\,\,p+q+r=0 \\ \text{or,}\,\,\,  p^2+q^2+r^2-pq-qr-rp=0 \\ \Rightarrow 2p^2+2q^2+2r^2-2pq-2qr-2rp=0 \\ \Rightarrow (p-q)^2+(q-r)^2+(r-p)^2=0 \cdots(6)$

From (6), we can say $\,\,(p-q)^2=(q-r)^2=(r-p)^2=0 \\ \Rightarrow p=q=r.\,\,\text{(proved)}$

27. If the two equations $\,x^2 + ax + b = 0\,$ and $\,x^2+bx+a=0\,\,(a \neq b)\,\,$  have a common root, show that other roots are the roots of the equation $\,\,x^2+x+ab=0.$

Sol. Let $\,\,\alpha\,,\beta\,$ be the roots of  $\,x^2 + ax + b = 0\,\cdots (1)$ and $\,\alpha,\delta\,$ be the roots of  $\,x^2+bx+a=0\,\,\cdots(2)$

So, $\,\,\alpha+\beta=-a \cdots(3) ,\,\,\alpha\beta=b \cdots(4),\\ \alpha+\delta=-b\cdots(5),\,\alpha\delta=a\cdots(6)$

Since $\,\alpha\,$ is the common root of (1) and (2), we have $\,\,\alpha^2+a\alpha+b=0 \\ \text{and}\,\,\alpha^2+b\alpha+a=0$

Now, by cross-multiplication, we get $\,\frac{\alpha^2}{a^2-b^2}=\frac{\alpha}{b-a}=\frac{1}{b-a} \\ \Rightarrow \alpha^2 =\frac{a^2-b^2}{b-a}\\~~~~~~~~~~~=-(a+b)\cdots(7),\\ \text{and}\,\,\alpha=\frac{b-a}{b-a}=1\cdots(8)$

Hence, from (7) and (8), we get $\,\,1^2=-(a+b)\\ \Rightarrow a+b+1=0$

Again, adding (3) and (5), we get $\,2+\beta+\delta=-(a+b)\,\,[\text{since},\,\,\alpha=1]\\ \Rightarrow \beta+\delta=-(-1)-2=-1 \\ [\text{since,}\,a+b=-1]$

Now, multiplying (4) and (6), we get $\,\,\beta \delta=ab \,\,[\alpha=1]$

Hence , the equation having roots $\,\beta, \delta\,$ is : $\,x^2-(\beta+\delta)x+\beta\delta=0 \\ \Rightarrow x^2+x+ab=0$

Hence  the result.  

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