Differentiation (Part-38) | S N De

  In the previous article , we have discussed  Long Answer Type Questions (From Qst. 1-7) of S.N. Dey Math Exercise. In this article,  we are going to start few more  Long Answer Type Questions of Quadratic Equations (Part-6) as a part of our S.N. Dey Math solution series.  So, let's start.

8. Express the roots of the equation $\,\,q^2x^2-(p^2-2q)x+1=0\,$ in terms of those of $\,x^2+px+q=0.$

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

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

9.If the roots of the quadratics $\,x^2+2px+q=0\,\,$ and $\,\,x^2+2px+p=0\,(p \neq q)\,$ differ by a constant , show that $\,\,p+q+1=0.$

Sol. Let $\,\,\alpha,\,\beta\,$ be the roots of $\,x^2+2px+q=0\,\,$ and $\,\,\gamma,\,\delta\,$ be the roots of $\,\,x^2+2px+p=0\,(p \neq q).$ 

So, $\,\, \alpha+\beta=-2p,\\ \alpha\beta=q, \\ \gamma+\delta=-2q , \\ \gamma\delta=p.$

By question, $\,\,|\alpha-\beta|=|\gamma-\delta| \\ \Rightarrow (\alpha-\beta)^2=(\gamma-\delta)^2 \\ \Rightarrow (\alpha+\beta)^2-4\alpha\beta=(\gamma+\delta)^2-4\gamma\delta \\ \Rightarrow 4p^2-4q=4q^2-4p \\ \Rightarrow p^2-q=q^2-p \\ \Rightarrow (p^2-q^2)+(p-q)=0 \\ \Rightarrow (p+q)(p-q)+(p-q)=0 \\ \Rightarrow (p-q)(p+q+1)=0 \\ \Rightarrow p+q+1=0\,\,\,[\text{Since,}\,\, p\neq q]$

10. If one root of the equation $\,\,px^2+qx+r=0\,$ is the square of the other , prove that $\,\,q^3+pr^2+p^2r=3pqr.$

Sol. Let $\,\, \alpha,\,\alpha^2\,$ be the roots of the equation $\,\,px^2+qx+r=0.$

So, $\,\,\alpha+\alpha^2=-\frac qp \cdots(1)\\ \alpha \times \alpha^2=\frac rp \\ \Rightarrow \alpha^3=\frac rp.\cdots(2)$

Hence from (1), we get , $\,\alpha(1+\alpha)=-\frac qp \\ \Rightarrow \alpha^3(1+\alpha)^3=-\frac{q^3}{p^3} \\ \Rightarrow \frac rp [1+\alpha^3+3\alpha(1+\alpha)]=-\frac{q^3}{p^3} \\ \Rightarrow \frac rp [1+\frac rp+3 (-\frac qp)]=-\frac{q^3}{p^3} \\ \Rightarrow \frac{r}{p^2}(p+r-3q)=-\frac{q^3}{p^3} \\ \Rightarrow rp(p+r-3q)=-q^3 \\ \Rightarrow rp^2+r^2p-3pqr=-q^3 \\ \Rightarrow q^3+pr^2+p^2r=3pqr.$

11. If one root of the equation $\,\,ax^2+bx+c=0\,\,$ is the cube of the other, prove that, $\,(b^2-2ca)^2=ca(c+a)^2.$

Sol. Let $\,\alpha,\,\alpha^3\,$ be the roots of the equation $\,\,ax^2+bx+c=0 \\ \therefore \alpha+\alpha^3=-\frac ba \\ \text{and}\,\, \alpha \times \alpha^3=\frac ca \\ \Rightarrow \alpha^4=\frac ca \\ \text{Now,}\,\,\alpha(1+\alpha^2)=-\frac ba \\ \Rightarrow \alpha^2(1+\alpha^2)^2=\frac{b^2}{a^2}\\ \Rightarrow \alpha^2(1+\alpha^4+2 \alpha^2)=\frac{b^2}{a^2}\\ \Rightarrow \alpha^2(1+\frac ca)+\frac{2c}{a}=\frac{b^2}{a^2}\,\,[\text{Since,}\,\,\alpha^4=\frac ca]\\ \Rightarrow \alpha^4(\frac{c+a}{a})^2=\left(\frac{b^2-2ca}{a^2}\right)^2 \\ \Rightarrow \frac ca. \frac{(c+a)^2}{a^2} =\frac{(b^2-2ca)^2}{a^4} \\ \Rightarrow (b^2-2ca)^2=ca(c+a)^2$

12. If the roots of the equation $\,\,ax^2+bx+c=0\,$ are two consecutive integers then prove that $\,\,b^2-a^2=4ac.$

Sol. Let the roots of the equation $\,\,ax^2+bx+c=0\,$ is $\,\alpha,\,\,\alpha+1 \\ \therefore \alpha+(\alpha+1)=-\frac ba \cdots(1) \\ \text{and}\,\,\alpha(\alpha+1)=\frac ca \cdots(2)$

From (1), we get $\,\,2\alpha+1=-\frac ba \\ \Rightarrow 2\alpha=-\frac ba-1 \\ \Rightarrow \alpha=-\left(\frac{a+b}{2a}\right) \\ \text{Now by (2), we get}\,\,-\left(\frac{a+b}{2a}\right)\left(-\frac{a+b}{2a}+1\right)\\=\frac ca \\ \Rightarrow \frac{b^2-a^2}{4a^2}=\frac ca \\ \Rightarrow b^2-a^2=4ac$

13. If $\,\alpha\,$ and $\,\beta\,$ be the roots of $\,\,ax^2+2bx+c=0\,$ and $\,\,\alpha+\delta,\, \beta+\delta\,$ be those of $\,Ax^2+2Bx+C=0,\,$ prove that , $\,\,\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$ 

Sol.  Since $\,\alpha\,$ and $\,\beta\,$ be the roots of $\,\,ax^2+2bx+c=0 \\ \therefore \alpha+\beta=-\frac{2b}{a} \\ \text{and}\,\,\alpha\beta=\frac ca$

Again,  $\,\,\alpha+\delta,\, \beta+\delta\,$ be those of $\,Ax^2+Bx+C=0 \\ \therefore \alpha+\delta+\beta+\delta=-\frac{2B}{A} \\ \Rightarrow -\frac{2b}{a}+2\delta=-\frac{2B}{A} \\ \delta=\frac ba-\frac BA \\ \text{and}\,\,\,(\alpha+\delta)(\beta+\delta)=\frac CA \\ \Rightarrow \alpha\beta+(\alpha+\beta)\delta+\delta^2=\frac CA \\ \Rightarrow \frac ca-\frac{2b}{a}(\frac ba-\frac BA)+\left(\frac ba-\frac BA\right)^2=\frac CA \\ \Rightarrow \frac ca-\frac{2b^2}{a^2}+\frac{2Bb}{Aa}+\frac{b^2}{a^2}+\frac{B^2}{A^2}-\frac{2Bb}{Aa}=\frac CA \\ \Rightarrow \frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}.$

14. If the ratio of the roots of the equation $\,x^2-2px+q^2=0\,$ be equal to the ratio of the roots of the equation $\,\,x^2-2rx+s^2=0,\,$ prove that , $\,\,p^2s^2=q^2r^2.$

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

Again, let $\,\gamma,\,\delta\,$ be  the roots of the equation $\,\,x^2-2rx+s^2=0 \\ \therefore \gamma+\delta=2r \\ \text{and}\,\,\gamma\delta=s^2$

Hence, according to the problem, $\,\,\frac{\alpha}{\beta}=\frac{\gamma}{\delta} \\ \Rightarrow \frac{\alpha+\beta}{\alpha-\beta}=\frac{\gamma+\delta}{\gamma-\delta} \\ [\text{By componendo-dividendo}] \\ \Rightarrow \left(\frac{\alpha+\beta}{\alpha-\beta}\right)^2=\left(\frac{\gamma+\delta}{\gamma-\delta}\right)^2 \\ \Rightarrow \frac{(\alpha+\beta)^2}{(\alpha+\beta)^2-4\alpha\beta}=\frac{(\gamma+\delta)^2}{(\gamma+\delta)^2-4\gamma\delta} \\ \Rightarrow \frac{4p^2}{4p^2-4q^2}=\frac{4r^2}{4r^2-4s^2} \\ \Rightarrow p^2r^2-p^2s^2=p^2r^2-q^2r^2 \\ \Rightarrow p^2s^2=q^2r^2.$ 

15.  If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,\,x^2+px+q=0,\,$ show that $\,\,\frac{\alpha}{\beta}\,$ is a root of the equation $\,\,qx^2-(p^2-2q)x+q=0.$

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

Now, $\,\,qx^2-(p^2-2q)x+q=0 \\ \Rightarrow \alpha\beta x^2-[(\alpha+\beta)^2-2\alpha\beta]x+q=0 \\ \Rightarrow \alpha\beta x^2-(\alpha^2+\beta^2)x-\alpha\beta=0 \\ \Rightarrow \alpha\beta x^2-\alpha^2 x-\beta^2 x-\alpha\beta=0 \\ \Rightarrow \alpha x(\beta x-\alpha)-\beta(\beta x-\alpha)=0 \\ \Rightarrow (\beta x-\alpha)(\alpha x -\beta)=0 \\ \therefore x=\frac{\alpha}{\beta},\,\,\frac{\beta}{\alpha}$

To continue with the next part of this article, click here. 

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

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