Differentiation (Part-38) | S N De

 7. If $\,\alpha\,$ be a root of the equation $\,\,ax^2+bx+c=0,\,\,$ show that $\,\,k\alpha\,(k \neq 0)\,\,$ is a root of the equation $\,\,ax^2+bkx+ck^2=0.$

Sol. Since $\,\alpha\,$ be a root of the equation $\,\,ax^2+bx+c=0, \\ \Rightarrow a\alpha^2+b\alpha+c=0 \cdots(1)$

Now, we see $\,\,a(k\alpha)^2+bk(k\alpha)+ck^2\\=k^2(a\alpha^2+b\alpha+c)\\=k^2 \times 0\,\,[\,\text{By (1)}]\\=0 \cdots(2)$

So, by (2) we notice that  the equation $\,\,ax^2+bkx+ck^2=0\,\,$ is satisfied by $\,\,x=k\alpha.\,\,$ Hence the result. 

8. Find the quadratic equation with real coefficients which has $\,2-3i\,$ as a root $\,(i=\sqrt{-1}).$

Sol. We know that in a quadratic equation with real coefficients , imaginary roots occur in conjugate pairs. So, if  $\,2-3i\,$ is a root of that equation, then $\,2+3i\,$ is another root. 

So, we have the required equation given by : $\,x^2-\text{(sum of the roots)}x+\text{product of the roots}=0 \\ \Rightarrow x^2-[(2+3i)+(2-3i)]x+(2+3i)(2-3i)=0 \\ \Rightarrow x^2-4x+[2^2-(3i)^2]=0 \\ \Rightarrow x^2-4x+13=0$

9. Form a quadratic equation with rational coefficients whose one root is $\,\,4+\sqrt{7}.$

Sol. If   $\,\,4+\sqrt{7}\,\,$ is one root  of the quadratic equation with rational coefficients , then $\,4-\sqrt{7}\,$ is another root.

So, we have the required equation given by : 

$\,x^2-\text{(sum of the roots)}x \\+\text{product of the roots}=0 \\ \Rightarrow  x^2-[(4+\sqrt{7})+(4-\sqrt{7})]x\\+(4+\sqrt{7})(4-\sqrt{7})=0 \\ \Rightarrow x^2-8x+[4^2-(\sqrt7)^2]=0 \\ \Rightarrow x^2-8x+9=0.$

10. If $\,\alpha,\,\beta\,$ are the roots of the equation $\,\,x(x-3)=4,\,\,$ find the value of $\,\alpha^2+\beta^2.$

Sol. Since $\,\alpha,\,\beta\,$ are the roots of the equation $\,\,x(x-3)=4 \,\,\text{or,}\,\,x^2-3x-4=0,\,\,$ we have $\,\,\alpha+\beta=3,\,\,\alpha\beta=-4 \cdots(1)$

So, $\,\,\alpha^2+\beta^2\\=(\alpha+\beta)^2-2 \alpha\beta\\=(3)^2-2 \times (-4)\\=9+8\\=17$

11. If $\,\,a, b, c \,$ are in G.P., prove that the roots of the equation $\,\,ax^2+2bx+c=0\,\,$ are equal.

Sol.  Since $\,\,a, b, c \,$ are in G.P., so $\,\,b^2=ac.$

Now, $\,\,ax^2+2bx+c=0 \cdots(1) \\ \Rightarrow x=\frac{-2b \pm \sqrt{4b^2-4ac}}{2a} \\ \text{so that}\,\,\,\,\alpha=\frac{-2b + \sqrt{4b^2-4ac}}{2a},\cdots(3)\\ \beta=\frac{-2b - \sqrt{4b^2-4ac}}{2a} \cdots(4)$ 

where $\,\,\alpha,\,\,\beta\,\,$ are the roots of (1). 

Now from (3) and (4), we get $\,\,\alpha=\frac{-2b+\sqrt{4b^2-4b^2}}{2a}=-\frac ba \\ \beta=\frac{-2b-\sqrt{4b^2-4b^2}}{2a}=-\frac ba\,\,[\text{Since}\,\,\,b^2=ac] $

12. Find $\,m,\,\,$ given that the difference of the roots of the equation $\,\,2x^2-12x+m+2=0\,\,$ is $\,2.$

Sol. Let $\,\,\alpha,\,\beta\,[\,\,\alpha>\beta]\,$ be the roots of the equation $\,\,2x^2-12x+m+2=0\,\,$ 

 $ \therefore\,\alpha+\beta=-\frac{-12}{2}=6\cdots(1) \\ \text{and}\,\,\,\alpha\beta= \frac{m+2}{2}\cdots(2)$

Again, by condition , we have $\,\,\,\alpha-\beta=2 \cdots(3)$

Hence, from (1) and (3), we get $\,\,(\alpha+\beta)+(\alpha-\beta)=6+2 \\ \Rightarrow 2\alpha=8 \\ \Rightarrow \alpha=\frac 82=4 \\ \therefore \beta=6-\alpha\,\,\,[\text{By (1)}]\\~~~~~~=6-4\\~~~~~~=2$

So, from (2) we get $\,\,4 \times 2=\frac{m+2}{2} \\ \Rightarrow 8 \times 2=m+2 \\ \Rightarrow m=14.$

13. If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,\,(x-a)(x-b)=c\,\,(c \neq 0)\,\,$, prove that $\,a\,$ and $\,b\,$ are the roots of the equation $\,\,(x-\alpha)(x-\beta)+c=0.$

Sol. We have the equation $\,\,(x-a)(x-b)=c\,\,(c \neq 0)\\ \Rightarrow x^2-(a+b)x+ab-c=0 \cdots(1)$ 

 Since $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation (1), we have $\,\,\alpha+\beta=-\frac{-(a+b)}{1}=a+b \cdots(2) \\ \text{and}\,\,\alpha\beta=ab-c\cdots(3)$

Now, $\,\,(x-\alpha)(x-\beta)+c=0 \\ \Rightarrow x^2-(\alpha+\beta)x+\alpha\beta+c=0 \\ \Rightarrow x^2-(a+b)x+(ab-c)+c=0 \\~~~~~~~~~~~~~~~~~~~~~~~~~~~ [\text{By (2) and (3)}]\\ \Rightarrow x^2-(a+b)x+ab=0 \\ \Rightarrow (x-a)(x-b)=0 \cdots(4)$

From (4), we can conclude that $\,a\,$ and $\,b\,$ are the roots of the equation $\,\,(x-\alpha)(x-\beta)+c=0.$

14. If $\,\,2+\sqrt3 i\,\,$ is a root of the equation $\,\,x^2+px+q=0,\,\,$ where $\,p\,$ and $\,q\,$ are real, find $\,p\,$ and $\,q.$

Sol. We know that in a quadratic equation with real coefficients , imaginary roots occur in conjugate pairs. So, if $\,\,2+\sqrt3 i\,\,$ is a root of that equation, then $\,\,2-\sqrt3 i\,\,$ is another root.

Now, we have the equation $\,\,x^2+px+q=0 \cdots(1),\,\,$ where $\,p\,$ and $\,q\,$ are real. 

So, sum of the roots of (1) is : $\,\,2+\sqrt3 i+2-\sqrt3 i=-p/1 \\ \Rightarrow 4=-p \\ \Rightarrow p=-4.$

Also, the product of the roots of (1) is : $\,\,(2+\sqrt3 i)(2-\sqrt3 i)=q/1 \\ \Rightarrow 2^2-(\sqrt3 i)^2=q \\ \Rightarrow 4+3=q\,\,\,\,[\text{Since}\,\,i^2=-1] \\ \Rightarrow q=7.$

15.  Find the value of $\,p\,$ so that the roots of the equation  $\,\,3x^2-2(7+9p)x+(8-5p)=0\,\,$ are reciprocal to one another.

Sol.  Since the roots of the equation  $\,\,3x^2-2(7+9p)x+(8-5p)=0\,\,$ are reciprocal to one another, let $\,\,\alpha\,$ and $\,\,\frac{1}{\alpha}\,\,$ be  two roots .

Now, by the given problem, $\,\, \alpha \times \frac{1}{\alpha}=\frac{8-5p}{3} \\ \Rightarrow 1=\frac{8-5p}{3} \\ \Rightarrow 3=8-5p \\ \Rightarrow 5p=8-3 \\ \Rightarrow p=\frac 55=1.$

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