For what value/s of constant 'p' for which the given quadratic have both roots as infinity. $(2p^3-13p^2+27p-18)x^2 + (2p^2-9p+9)x +2p^2-7p+6=0$ Options are :- $1) 3/2 2) 2 3) 3 4) /phi $

Since both roots are infinite then sum of the roots must be infinity. For this quadratic let alpha and beta be the roots then we say that Alpha + beta (sum of roots) = -(2p^2-9p+9)/(2p^3-13p^2+27p-18) For the sum to be infinity 2p^3-13p^2+27p-18 must equal to zero. On solving the equation 2p^3-13p^2+27p-18=0 we get p1 =3/2 p2=2 and p3 =3. For p1 and p3 2p^2-9p+9 become zero. So correct option for this question may be 2 but answer in my book is given as option 3. Why this is so.. plz explain me.


closed as unclear what you're asking by José Carlos Santos, Cesareo, mrtaurho, Paul Frost, GNUSupporter 8964民主女神 地下教會 Feb 15 at 13:33

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What do you mean by root at infinity ? $\endgroup$ – Jean Marie Feb 14 at 17:58
  • $\begingroup$ Roots as infinity means value of both roots to be infinity ( I think so) $\endgroup$ – saket kumar Feb 14 at 18:06
  • $\begingroup$ Is this $$(2p^3-13p^2+27p-18)x^2+(2p^2-9p+9)x+2p^2-7p+6=0$$? $\endgroup$ – Dr. Sonnhard Graubner Feb 14 at 18:09
  • $\begingroup$ Yes @ Dr. Sonnhard $\endgroup$ – saket kumar Feb 14 at 18:11
  • $\begingroup$ See in my solution the convention that a root is at infinity iff its inverse is zero. Have you already studied derivatives ? $\endgroup$ – Jean Marie Feb 14 at 18:11

Instead of using the term "infinite $x$", we are going to invert things by setting $x=\frac{1}{y}$. And say by definition that "$x$ infinite" means that its inverse $y$ is zero.

But we must be a little cautious. Let us, in the initial equation, $(2p^3-13p^2+27p-18)x^2 + (2p^2-9p+9)x +2p^2-7p+6=0$

replace first $x$ by $\frac{1}{y}$ :

$$\underbrace{(2p^3-13p^2+27p-18)}_A\tfrac{1}{y^2} + \underbrace{(2p^2-9p+9)}_B\tfrac{1}{y} +\underbrace{(2p^2-7p+6)}_C=0 \tag{1}$$

Reducing the LHS (Left Hand Side) to a same denominator $y^2$, this equation is converted into "numerator = 0" which means a quadratic equation in variable $y$ :

$$\underbrace{(2p^2-7p+6)}_Cy^2+\underbrace{(2p^2-9p+9)}_By+\underbrace{(2p^3-13p^2+27p-18)}_A=0 \tag{2}$$

(note that the order of coefficients has been reversed between (1) and (2)).

This equation $Cy^2+By+A=0$ has a double root in $0$ iff it is of the form $Cy^2=0$. Thus it is equivalent to say that coefficients $B$ and $A$ are $0$, thus verify the system

$$\begin{cases}B&=&2p^3-13p^2+27p-18&=&0 &\ \ (b)\\ A&=&2p^2-9p+9&=&0 &\ \ (a)\end{cases}$$

It remains for you to solve the system (a) and (b) .

Hint : multiply (a) by $p$ and substract to (b) : you will get a quadratic in $p$....

  • $\begingroup$ For roots to be infinity, Can we say that the coffecient of X^2 and coefficient of X must be equal to zero and Constant term must not be equal to zero. $\endgroup$ – saket kumar Feb 14 at 18:22
  • 1
    $\begingroup$ @Saket Let the quadratic equation be $ax^2+bx+c=0$. Divide by $x^2$ on both sides to get$$a+b/x+c/x^2=0$$Since $\infty$ is a double root, this equation must be of the form $k\cdot\frac1x\cdot\frac1x,k\in\Bbb R-\{0\}$. This means $a=b=0,c\ne0$. $\endgroup$ – Shubham Johri Feb 14 at 18:28
  • $\begingroup$ How you write k.1/x.1/x explain plz.. $\endgroup$ – saket kumar Feb 14 at 18:35
  • $\begingroup$ @saketkumar A quadratic with roots $p,q$ is written as $a(x-p)(x-q)$. The equation in $1/x$ has both roots $0$, so it will be written as $k(1/x-0)(1/x-0)$ $\endgroup$ – Shubham Johri Feb 14 at 18:42
  • 1
    $\begingroup$ I have modified my answer in order (using the simpler approach suggested by @Shubham Johri): no need in fac to use derivatives. $\endgroup$ – Jean Marie Feb 14 at 18:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.