Differentiation (Part-38) | S N De

 

28. If $\,c\,$ is real, show that the roots of the quadratic equation $\,cx^2+(c-1)x+1-2c=0\,\,$ are real. If the sum of the roots of the equation be equal to three times their difference , then find $\,\,c.$

Sol. Let  the discriminant of quadratic equation $\,cx^2+(c-1)x+1-2c=0\cdots(1)$ is $\,D.$

So, $\,D=(c-1)^2-4.c.(1-2c)\\~~~~=(c-1)^2-4c+8c^2\\~~~~=9c^2-6c+1\\~~~~=(3c-1)^2\geq 0$

Since the roots of (1) are real and $\,\,D \geq 0,\,\,$ so the roots of (1) are real. (proved)

Let $\,\,\alpha,\beta\,$ be the roots of (1). 

Then $\,\alpha+\beta=-\frac{c-1}{c},\,\, \alpha\beta=\left(\frac{1-2c}{c}\right)$

By question, we have $\alpha+\beta=3 \alpha\beta \\ \Rightarrow -\frac{c-1}{c}=\frac{3(1-2c)}{c} \\ \Rightarrow -c+1=3-6c \,\,[c\neq 0]\\ \Rightarrow c=\frac 25.$ 

29. If $\,\alpha\,$ is a root of the equation $\,\,ax^2+bx+c=0\,$ then show that $\,m\alpha^2\,(m \neq 0)\,\,$ is a root of the equation $\,a^2x^2+(2ac-b^2)mx+m^2c^2=0.$

Sol.  Since $\,\alpha\,$ is a root of the equation $\,\,ax^2+bx+c=0 \\ \therefore a\alpha^2+b\alpha+c=0$ 

Now, we put $\,\,x=m\alpha^2,\,\,c=-a\alpha^2-b\alpha\,\,$ in the equation $\,a^2x^2+(2ac-b^2)mx+m^2c^2=0\cdots(1)$

So, $\,a^2m^2\alpha^4+[2a(-a\alpha^2-b\alpha)-b^2]m^2\alpha^2\\+m^2[-(a\alpha^2+b\alpha)]^2\\=a^2m^2\alpha^4-2a^2m^2\alpha^4-2abm^2\alpha^3-b^2m^2\alpha^2\\+m^2(a^2\alpha^4+2ab\alpha^3+b^2\alpha^2)\\=a^2m^2\alpha^4-2a^2m^2\alpha^4-2abm^2\alpha^3-b^2m^2\alpha^2\\+a^2m^2\alpha^4+2abm^2\alpha^3+b^2m^2\alpha^2\\=0.$

Hence the equation (1) is satisfied by $\,\,x=m\alpha^2\,\,$ and so we can conclude that $\,m\alpha^2\,(m \neq 0)\,\,$ is a root of the equation $\,a^2x^2+(2ac-b^2)mx+m^2c^2=0.$

30. If the quadratic equations $\,x^2+ax+b=0\,\,$ and $\,x^2+bx+a=0\,(a \neq b)\,$ have a common root, find the numerical value of $\,(a+b).$

Sol. If the quadratic equations $\,x^2+ax+b=0\,\,$ and $\,x^2+bx+a=0\,(a \neq b)\,$ have a common root $\,\,\alpha\,\,$ then $\,\alpha^2+a\alpha+b=0 \cdots(1) \\  \alpha^2+b\alpha+a=0 \cdots(2)$

By cross-multiplication  we get from (1) and (2), 

$\,\frac{\alpha^2}{a^2-b^2}=\frac{\alpha}{b-a}=\frac{1}{b-a} \\ \therefore \alpha^2=-\frac{a^2-b^2}{a-b}\\~~~~~~~~=-\frac{(a+b)(a-b)}{a-b}\\~~~~~~~~=-(a+b)\\ \text{and}\,\,\, \alpha=1 \\ \therefore a+b=-1 \\ \Rightarrow |a+b|=1$

Hence,  the numerical value of $\,(a+b)=1.$

31. If the roots of the equation $\,\,x^2-px+q=0\,\,$ be $\,\alpha,\beta\,$ and the roots of the equation $\,\,x^2-ax+b=0\,$ be $\,\alpha,\frac{1}{\beta}\,$ then prove that, $\,bq(p-a)^2=(q-b)^2$

Sol. If the roots of the equation $\,\,x^2-px+q=0\,\,$ be $\,\alpha,\beta\,$ , then $\,\,\alpha+\beta=p \cdots(1)\,\,\,\,\,\alpha+\beta=q\cdots(2)$

Again, if the roots of the equation $\,\,x^2-ax+b=0\,$ be $\,\alpha,\frac{1}{\beta}\,$ then $\,\,\alpha+\frac{1}{\beta}=a \cdots(3),\,\,\frac{\alpha}{\beta}=b \cdots(4)$

Now, from (1) and (3), we get $\,\,(\beta-\frac{1}{\beta})=p-a \cdots(5) $

Again, from (2) and (4), we get, $\,\,\alpha(\beta-\frac{1}{\beta})=q-b \cdots(6)$

Now, multiplying (2) and (4), we get $\,\,\alpha^2=bq \cdots(7)$

Also, squaring both sides of (6), we get $\,\,\alpha^2(\beta-\frac{1}{\beta})^2=(q-b)^2 \\ \Rightarrow bq(p-a)^2=(q-b)^2\,\,[\text{By (5)}]$

32. If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,5x^2+bx+c=0\,$ show that $\,5x^2+bx+c=5(x-\alpha)(x-\beta)\,$ 

Show also that for all real values of $\,x\,$ the expression $\,5x^2 + bx+ c\,$ cannot be negative if (i) $\,\alpha\,$ and $\,\beta\,$ are real and equal or (ii) $\,\alpha\,$ and $\,\beta\,$ are conjugate complex.

Sol.    If $\,\alpha\,$ and $\,\beta\,$ be the roots of the equation $\,5x^2+bx+c=0\,$  then $\,\alpha+\beta=-\frac b5,\,\,\alpha\beta=\frac c5.$

Now, $\,\,5x^2+bx+c\\=5\left(x^2+\frac b5 x+\frac c5\right)\\=5\left[x^2-(\alpha+\beta)x+\alpha\beta\right]\\=5[x^2-\alpha x-\beta x+\alpha\beta]\\=5(x-\alpha)(x-\beta)$

Again, $\,\,5x^2+bx+c\\=5[x^2-(\alpha+\beta)x+\alpha\beta]\\=5\left[x^2-2.x.(\frac{\alpha+\beta}{2})+(\frac{\alpha+\beta}{2})^2+\alpha\beta\\-(\frac{\alpha+\beta}{2})^2\right]\\=5\left[(x-\frac{\alpha+\beta}{2})^2-(\frac{\alpha-\beta}{2})^2\right]$

Hence, (i) if $\,\alpha,\, \beta\,$ are real and equal, then $\,5x^2+bx+c \geq 0.$

(ii) If $\,\alpha\,$ and $\,\beta\,$ are conjugate complex,  $\,\,5x^2+bx+c \geq 0.$

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