Question

Solve this -
Q. If p and q are real and , $p\ne q$ then show that the roots of the equation $\left(p-q\right)x^2+5\left(p+q\right)x-2(p-q)=0$ are real and unequal.

Answer 1

Dear Student,

$$\begin{gathered} Aquadraticequationa{x}^2+bx+c=0hasrealandunequalrootsif{b}^2-4ac>0. \\ Givenequation:\left(p-q\right)x^2+5\left(p+q\right)x-2\left(p-q\right)=0 \\ Here,{\left[5\left(p+q\right)\right]}^2-4\left(p-q\right)\left(-2\left(p-q\right)\right) \\ =25{\left(p+q\right)}^2+8{\left(p-q\right)}^2,whichisthesumoftwosquarenumbers. \\ Thesumoftwosquarenumbersisalwayspositiveorequalto0. \\ Now,25{\left(p+q\right)}^2+8{\left(p-q\right)}^2=0 \\ \Rightarrow {\left(p+q\right)}^2=0and{\left(p-q\right)}^2=0 \\ \Rightarrow p=-qandp=q,whichisnotpossible. \\ So,25{\left(p+q\right)}^2+8{\left(p-q\right)}^2>0. \\ Hence,thegivenequationhasrealandunequalroots. \end{gathered}$$