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.
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 $p$....