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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.

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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

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  • $\begingroup$ 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
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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 $p$....

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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

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