Question

If $p,q$ are real and $p \ne q$, then the roots of the equation $(p - q){x^2} + 5(p + q)x - 2(p - q) = 0$ are real and unequal.

Answer 1

Hint:- In this question first we need to compare the coefficients of given quadratic equation$(p - q){x^2} + 5(p + q)x - 2(p - q) = 0$ with standard form quadratic $a{x^2} + bx + c = 0$. And then find the discriminant if it is greater than zero then roots will be real and unequal otherwise they may be equal or imaginary.

Complete step-by-step solution -
Given quadratic equation is $(p - q){x^2} + 5(p + q)x - 2(p - q) = 0$
Standard form quadratic equation is $a{x^2} + bx + c = 0$ and its discriminant is given by
$D = {b^2} - 4ac$ ………………………... eq.1
Now, for real and unequal roots discriminant must be greater zero $(D > 0)$ because only in this condition $\dfrac{{ - b \pm D}}{2a}$ gives two different values which are roots of $a{x^2} + bx + c = 0$.
On comparing given quadratic equation with standard quadratic form equation, we get
$ \Rightarrow a = p - q;{\text{ }}b = 5(p + q);{\text{ c = }} - {\text{2(}}p - q{\text{)}}$
Put $a,b,c$ in eq.1 to get the discriminant
$
   \Rightarrow D = {[5(p + q)]^2} - 4(p - q) \times \{ - 2(p - q)\} \\
   \Rightarrow D = 25{(p + q)^2} + 8{(p - q)^2} \\
 $
Since, ${(p + q)^2}{\text{ and }}{(p - q)^2}$ are the perfect squares and their values will always be greater than zero because it is given that $p \ne q$.
Hence, the roots of a given quadratic equation $(p - q){x^2} + 5(p + q)x - 2(p - q) = 0$ are real and unequal.

Note:- Whenever you get this type of question the key concept to solve this is to find the discriminant of the given quadratic equation and check it with the given information. And remember one thing that discriminant reveals what type of roots of the given quadratic equation has whether it has two solutions, one solution or no solution.