Differentiation (Part-38) | S N De

 

 In the previous article , we have discussed Long Answer Type Questions (From Q. 8-15) of S.N. Dey Math Exercise. In this article,  we are going to start Long Answer Type Questions of Quadratic Equations (Part-6) as a part of our S.N. Dey Math solution series.  So, let's start.

16. If $\,\,p^3-q(3p-1)+q^2=0,\,$ find the relation between the roots of the equation $\,\,x^2+px+q=0.$

Sol. Let $\,\,\alpha,\beta\,\,$ be the roots  of the equation $\,\,x^2+px+q=0 \cdots(1) \\ \text{so that}\,\,\alpha+\beta=-p ,\\ \alpha\beta=q.$

Now, $\,\,p^3-q(3p-1)+q^2=0 \\ \Rightarrow [-(\alpha+\beta)]^3-\alpha\beta[-3(\alpha+\beta)-1]\\+\alpha^2\beta^2=0 \\ \Rightarrow (\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)-\alpha\beta-\alpha^2\beta^2=0 \\ \Rightarrow \alpha^3+\beta^3-\alpha\beta-\alpha^2\beta^2=0 \\ \Rightarrow \alpha(\alpha^2-\beta)-\beta^2(\alpha^2-\beta)=0 \\ \Rightarrow (\alpha^2-\beta)(\alpha-\beta^2)=0 \\ \therefore \alpha^2=\beta,\,\,\text{or}\,\, \alpha=\beta^2.$

Hence, one root of (1) will be square of another root.

17. Let $\,\,a,b,c\,$ be real numbers with $\,\,a \neq 0\,$ and let $\,\,\alpha,\,\beta\,\,$ be the roots of the equation $\,\,ax^2+bx+c=0.\,\,$ Express the roots of $\,\,a^3x^2+abcx+c^3=0\,\,$ in terms of $\,\,\alpha,\,\,\beta.$

Sol. If $\,\,\alpha,\,\beta\,\,$ be the roots of the equation $\,\,ax^2+bx+c=0 \\ \text{then,}\,\,\alpha+\beta=-\frac ba \\ \text{and}\,\,\alpha \beta=\frac ca.$

Now,  $\,\,a^3x^2+abcx+c^3=0 \\ \Rightarrow x^2+\frac ba.\frac ca x+\left(\frac ca \right)^3=0 \\ \Rightarrow x^2-(\alpha+\beta)\alpha\beta x+\alpha^3\beta^3=0 \\ \Rightarrow x^2-\alpha^2\beta x-\alpha \beta^2 x+\alpha^3 \beta^3=0 \\ \Rightarrow x(x-\alpha^2 \beta)-\alpha \beta^2(x-\alpha^2 \beta)=0 \\ \Rightarrow (x-\alpha^2 \beta)(x-\alpha \beta^2)=0 \\ \Rightarrow x=\alpha^2 \beta,\,\,\, \alpha \beta^2.$

Hence, the roots of the equation $\,\,a^3x^2+abcx+c^3=0\,\,$ is $\,\,\alpha^2\beta,\,\,\alpha\beta^2.$

18. If $\,\alpha\,$ be a root of the quadratic equation $\,4x^2+2x-1=0,\,$ prove that $\,4 \alpha^3-3\alpha\,$ is the other root.

Sol. If $\,\alpha\,$ be a root of the quadratic equation $\,4x^2+2x-1=0 \cdots(1),\,$ let $\,\beta\,$ be another root . Then $\,\alpha+\beta=-\frac 24=-\frac 12 \cdots (1A) \\ \text{and}\,\,\alpha\beta=-\frac 14.$

By (1), we have $\quad 4\alpha^2+2\alpha-1=0 \\ \Rightarrow 4\alpha^2=1-2\alpha \cdots(2)$

Now $\,\,4\alpha^3-3 \alpha\\=\alpha \times 4\alpha^2-3\alpha\\=\alpha(1-2\alpha)-3 \alpha \,\,[\text{By (2)}]\\=\alpha-2\alpha^2-3\alpha\\=-2\alpha-\frac 12\times 4\alpha^2\\=-2\alpha-\frac 12(1-2\alpha)\\=-2\alpha-\frac 12+\alpha\\=-\frac 12-\alpha\\=\beta \,\,\,[\text{By (1A)}]$

Hence, follows the result.

19.  Let $\,\,\alpha,\,\beta\,$ be the roots of the equation $\,\,x^2-3x+a=0\,$and $\,\,\gamma,\,\delta\,$ the roots of the equation $\,\,x^2-12x+b=0.$ If $\,\,\alpha,\beta,\gamma,\delta\,$ are in G.P. having positive common ratio, then find the values of $\,a\,$ and $\,b.$

Sol. Since $\,\,\alpha,\,\beta\,$ be the roots of the equation $\,\,x^2-3x+a=0 ,\\ \alpha+\beta=3 \cdots (1) \\ \text{and}\,\, \alpha \beta=a \cdots(2)$

Again, since $\,\,\gamma,\,\delta\,$ the roots of the equation $\,\,x^2-12x+b=0, \\ \gamma+\delta=12 \cdots(3) \\ \text{and}\,\,\gamma\delta=b \cdots(4)$

But since $\,\,\alpha,\beta,\gamma,\delta\,$ are in G.P. having positive common ratio, say $(r)$

Therefore $\,\frac{\alpha}{\beta}=\frac{\beta}{\gamma}=\frac{\gamma}{\delta}=r \cdots(5) \\ \Rightarrow \gamma=r\delta,\,\beta=r \gamma=r^2\delta,\\ \text{and}\,\, \alpha=r\beta=r^3\delta$ 

Now, $\,\,\frac{\alpha+\beta}{\gamma+\delta}=\frac{3}{12} \\ \Rightarrow \frac{r^3\delta+r^2\delta}{r\delta+\delta}=\frac 14 \\ \Rightarrow r^2= \frac{1}{4} \\ \Rightarrow r=\pm \frac 12.$ 

Now without any loss of generality, let $\,\,r=\frac 12.$

Then by (5), we get $\,\,\frac{\alpha}{\beta}=\frac 12 \\ \Rightarrow \beta=2\alpha.$

Hence by (1), $\,\,\alpha+2\alpha=3 \\ \Rightarrow \alpha=\frac 33=1 \\ \text{and}\,\,\beta=3-1=2. \\ \text{So by (2)},\,\, a= 1\times 2=2 \\ \text{Again,}\,\,\alpha=r^3\delta \\ \Rightarrow 1=\frac 18 \times \delta \\ \Rightarrow \delta=8. \\ \text{Again,}\,\, \beta=r\gamma \\ \Rightarrow 2=\frac 12 \times \gamma \\ \Rightarrow \gamma=2\times 2=4.\\ \text{Hence},\,\,b=\gamma \delta= 4 \times 8=32.$ 

Similarly, by putting $\,\,r=-\frac 12 \,\text{we get}, \,\, r^3\delta+r^2\delta=3 \\ \Rightarrow -\frac 18 \delta+ \frac 14 \delta=3 \\ \Rightarrow \delta=24 \\ \text{and so,}\,\, a=\alpha\beta=r^5 \delta^2=\frac{24^2}{-2^5}=-18 \\ b=\gamma \delta=r \delta^2=\frac{24^2}{-2}=-288.$

20. If $\,\,a,b,c\,$ are real, show that the roots of each of the following equations are real :

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

Sol. (i) $\,(x-a)(x-b)=b^2 \\ \Rightarrow x^2-(a+b)x+(ab-b^2)=0 \cdots(1)$

Now, the discriminant of (i) is given by : $\,\,D=[-(a+b)]^2-4(ab-b^2)\\~~~~~=(a+b)^2-4ab+4b^2\\~~~~~=(a-b)^2+4b^2 \geq 0 \\~~~~~~~~~~~~~~~~~~~~~~ [\,\text{Since,}\,a,b,c \in \mathbb R ]$

Hence, the roots of (1) are real.

Sol. (ii)  $(b-c)x^2+2(c-a)x+a-b=0 \cdots(1)$

Now, the discriminant of (i) is given by : $\,\,D=[2(c-a)]^2-4(b-c)(a-b)\\~~~~~=4[(c-a)^2-(b-c)(a-b)]\\=4[c^2-2ca+a^2-ab+b^2+ac-bc]\\=2[2a^2+2b^2+2c^2-2ab-2bc-2ca]\\=2[(a-b)^2+(b-c)^2+(c-a)^2] \geq 0$

Hence, the roots of (1) are real.

20. If $\,\,a,b,c\,$ are real, show that the roots of each of the following equations are real :

(iii) $\,\,\frac{1}{x-p}+\frac{1}{x-q}=\frac{1}{a^2} \\ (iv)\,\frac{1}{x-a}+\frac{1}{x-1}+\frac{1}{x-2}=0$

Sol.  (iii) $\,\,\frac{1}{x-p}+\frac{1}{x-q}=\frac{1}{a^2} \\ \Rightarrow \frac{x-q+x-p}{(x-p)(x-q)}=\frac{1}{a^2} \\ \Rightarrow x^2-(p+q)x+pq=2a^2x-a^2(p+q) \\ \Rightarrow x^2-(p+q+2a^2)x+pq+a^2(p+q)=0 \cdots(1)$

Now, the discriminant of (1) is : $\,\,D=[-(p+q+2a^2)]^2-4[pq+a^2(p+q)]\\=(p+q+2a^2)^2-4pq-4a^2(p+q)\\=(p+q)^2+4a^2(p+q)+4a^4\\-4pq-4a^2(p+q)\\=(p-q)^2+(2a^2)^2 \geq 0$

Hence, the roots of (1) are real.

Sol. (iv) $\,\,\frac{1}{x-a}+\frac{1}{x-1}+\frac{1}{x-2}=0 \\ \Rightarrow (x-1)(x-2)+(x-a)(x-2)+(x-a)(x-1)=0 \\ \Rightarrow x^2-3x+2+x^2-(a+2)x+2a+x^2-(a+1)x+a=0 \\ \Rightarrow 3x^2-2(a+3)+(3a+2)=0 \cdots(1)$

The discriminant of (1) is : $\,\,D=[-2(a+3)]^2-4\times 3\times (3a+2)\\=4[(a+3)^2-3(3a+2)]\\=4(a^2-3a+3)\\=4\left[\left(a-\frac 32\right)^2+\frac 34\right]>0\,\,[a \in \mathbb R]$

Hence, the roots of (1) are real.