If the roots of the equation ({p^2} + {q^2}){ | vedclass

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(B) The given quadratic equation is $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$.Since the roots are real and equal,the discriminant $D =

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$G.P.$

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(B) The given quadratic equation is $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$.Since the roots are real and equal,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = ({p^2} + {q^2})$,$b = -2q(p + r)$,and $c = ({q^2} + {r^2})$.So,$D = [-2q(p + r)]^2 - 4({p^2} + {q^2})({q^2} + {r^2}) = 0$.$4q^2(p + r)^2 - 4({p^2}q^2 + p^2r^2 + q^4 + q^2r^2) = 0$.Dividing by $4$,we get $q^2(p^2 + r^2 + 2pr) - (p^2q^2 + p^2r^2 + q^4 + q^2r^2) = 0$.$p^2q^2 + q^2r^2 + 2pq^2r - p^2q^2 - p^2r^2 - q^4 - q^2r^2 = 0$.Simplifying the expression,we get $2pq^2r - p^2r^2 - q^4 = 0$.$-(q^4 - 2pq^2r + p^2r^2) = 0$.$-(q^2 - pr)^2 = 0$.This implies $q^2 - pr = 0$,or $q^2 = pr$.Therefore,$p, q, r$ are in $G.P.$

If the roots of the equation $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$ are real and equal,then $p, q, r$ will be in:

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