If `p ,q` are real `p!=q ,` then show that the roots of the equation `(p-q)x^2+5(p+q)x-2(p-q)=0` are real and unequal.

Explanation

Given

`(p-q)x^2+5(p+q)x-2(p-q)=0`

Here,

`a=p−q,b=5(p+q),c=−2(p−q)`

Discriminant=`b^2−4ac`

=`[5(p+q)]^2−4(p−q)(p−q)`

=`25(p+q)^2+8(p−q)^2`

as,`(p+q)^2` and `(p−q)^2` are perfect squares and their values will always be greater than 0.

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