If p, q and r are real, p≠q, then show that root of the equation (p-q)x2 +5(p+q)x-2(p-q) =0 are real and unequal
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D=b^2-4ac
=(p+q)^2-4*5(p-q)*(p-q)*(-2)
= (p+q)^2+40*(p-q)^2
square of a no cannot be negative
∴ (p+q)^2 and (p-q)^2 are >=0
but (p+q) and (p-q) both can never be zero
∴either of the two is zero or both are non-zero
and addition of position no.is always positive
∴D>0
∴ roots are real and roots are not equal(as D≠0)
=(p+q)^2-4*5(p-q)*(p-q)*(-2)
= (p+q)^2+40*(p-q)^2
square of a no cannot be negative
∴ (p+q)^2 and (p-q)^2 are >=0
but (p+q) and (p-q) both can never be zero
∴either of the two is zero or both are non-zero
and addition of position no.is always positive
∴D>0
∴ roots are real and roots are not equal(as D≠0)
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