Differentiation (Part-38) | S N De

 In previous article, we have discussed  Quadratic Equation : Long Answer Type Questions (33-38) of S.N. Dey math exercise .  In this article we will continue our journey to solve Very Short Answer Type Questions.

1. Find the condition for which the roots of the equation $\,\,ax^2-(a+1)x+1=0\,\,$ are always real.

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

Now, if the roots of (1) are to be real, then we must have the discriminant of (1) : $\,\,D \geq 0.$ Now, we compute $\,\,[-(a+1)]^2-4\times a \times 1 \\=(a+1)^2-4a\\=a^2+2a+1-4a\\=a^2-2a+1\\=(a-1)^2 \geq 0\,\,\text{only if a is real.}$

2. One root of the equation $\,\,3x^2-5x+c=0\,\,$ is $\,2;\,\,$ find its other root.

Sol. Since  one root of the equation $\,\,3x^2-5x+c=0\,\,$ is $\,2;\,\,$ we have , $\,3 \times 2^2-5 \times 2+c=0 \\ \Rightarrow 12-10+c=0 \\ \Rightarrow 2+c=0 \\ \Rightarrow c=-2$

Hence the given equation turns out to be $\,\,3x^2-5x-2=0 \cdots(1)$

Let $\,\,\alpha\,$ be the other root of (1) so that $\,\,\alpha+2=-(-5)/3 \\ \Rightarrow \alpha=\frac 53-2\\~~~~~~~~=-\frac 13.$

3. The product of the roots of the equation $\,\,3x^2+mx-(2m+3)=0\,\,$ is $\,5.\,$ Find $\,\,m.$

Sol. Since the product of the roots of the equation $\,\,3x^2+mx-(2m+3)=0\,\,$ is $\,5,\,$ 

we have $\,\,\frac{-(2m+3)}{3}=5 \\ \Rightarrow -(2m+3)=15 \\ \Rightarrow-2m-3=15 \\ \Rightarrow -2m=15+3 \\ \Rightarrow m=-\frac{18}{2}=-9.$

4. If one root of the equation $\,\,2x^2-5x+k=0\,\,$ be twice the other, find the value of $\,k.$

Sol. Let $\,\alpha,\,\,2\alpha\,\,$ be two roots of the aforesaid equation so that we have, 

$\,\,\alpha+2\alpha=-\frac{-5}{2} \\ \Rightarrow \alpha=\frac 56. \\ \text{Now,}\, \alpha \times 2 \alpha= \frac k2\\ \Rightarrow 2 \alpha^2=\frac k2 \\ \Rightarrow 2(\frac 56)^2=\frac k2 \\ \Rightarrow \frac{50}{36}=\frac k2 \\ \Rightarrow k=\frac{25}{9}.$

5. Find the condition so that the roots of the equation $\,\,lx^2+mx+n=0\,\,$ are (i) equal in magnitude and opposite in signs (ii) reciprocal to one another.

Sol. (i) Keeping in mind that the roots of the equation are  equal in magnitude and opposite in signs ,  let $\,\, \alpha,\,\, -\alpha\,\,$ be the roots of the aforesaid equation so that we have $\,\, \alpha+ (-\alpha)=-\frac ml \\ \Rightarrow 0=-\frac ml \\ \Rightarrow m=0.$

Sol. (ii) Since the roots of the equation are reciprocal to one another, we have $\,\, \alpha \times \frac{1}{\alpha}=\frac nl \\ \Rightarrow 1=\frac nl \\ \Rightarrow n=l.$

6. (i) If the ratio of the roots of the equation $\,\,x^2-px+q=0\,\,$ be $\,1:2,\,$ find the relation between $\,p\,$ and $\,\,q.$

Sol. Let $\,\alpha,\,2\alpha \,\,$ be the roots of the given equation so that $\,\,\alpha+2\alpha=-(-p)/1 \\ \Rightarrow 3\alpha=p \\ \Rightarrow \alpha=\frac p3 \\ \text{Again,}\,\,\alpha \times 2\alpha=q \\ \Rightarrow 2\alpha^2=q \\ \Rightarrow2 (\frac p3)^2=q \\ \Rightarrow 2p^2=9q.$

6(ii) If the roots of the equation $\,\,qx^2+px+q=0\,\,$ be imaginary, where $\,p,q >0,\,$ then show that , $\,\,0<p<2q.$

Sol. Since  ratio of the two roots of the given quadratic equation being imaginary then the discriminant must be $\, <0 \\ \text{which means}\,\,p^2-4q^2<0 \\ \Rightarrow p^2-(2q)^2 <0 \\ \Rightarrow (p+2q)(p-2q) <0 \cdots(1)$

But we are given $\,\,p,q >0 \Rightarrow p+2q >0 $

Hence by (1), we get $\,\, p-2q<0 \\ \Rightarrow p<2q \\ \Rightarrow 0<p<2q \,\,[\text{Since}\,\,p>0]$

6(iii) Find the condition for which the quadratic equation $\,\,ax^2+bx+c=0\,\,$ has exactly one zero root.

Sol. We have the quadratic equation $\,\,ax^2+bx+c=0 \\ \Rightarrow x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \cdots(1)$

By the given condition, we have $\,\,\sqrt{b^2-4ac}=b \\ \Rightarrow b^2-4ac=b^2 \\ \Rightarrow ac=0 \\ \Rightarrow c=0\,\,[\,\,\,\text{Since,}\,\, a\neq 0 \,\,\text{as the equation being quadratic}]$

Hence, the condition for which the quadratic equation $\,\,ax^2+bx+c=0\,\,$ has exactly one zero root is : $\,\,c=0,\,\,a\neq 0, b\neq 0.$

6(iv) If in the equation $\,\,4x^2+2bx+c=0,\,b=0,\,$ find the relation between the roots of the equation.

Sol.  We have the equation $\,\,4x^2+2bx+c=0,\,b=0 \\ \Rightarrow 4x^2+c=0 \\ \Rightarrow 4x^2=-c \\ \Rightarrow x^2=-\frac c4 \\ \Rightarrow x=\pm \sqrt{-\frac c4} \cdots(1)$

Hence, by (1) we can conclude that the roots of the equation are equal in magnitude and opposite in sign.