A-8. If a B are roots of x-px + 9 =...
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A-8. If a B are roots of x-px + 9 = 0 and a -2, B + 2 are roots of x-px + r = 0, then prove that 169 + (r + 4-9) = 4p2.

JEE/Engineering Exams
Maths
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( n^{2} cdot p n+2= ) ( arg 2=frac{p}{2} p=frac{-(b)}{c} ) ( n^{2}-p_{n}+9= ) ( dot{x}-2+beta+2=P ) ( alpha+beta=P ) ( (2-2)(3+2)=9 ) ( alpha beta+2 alpha-2 beta-4=21 ) ( 2(2+beta)=5+4 cdots alpha beta ) ( 2(2 beta)=9+4-2 ) ive have ( 162+(9+4-a)=4 p ) ( frac{operatorname{tg} 5}{16 q}+(2(2-B))^{2} ) ( -16 q+4(2-3)^{2} ) ( =162+40+left(2^{2}+8^{2}-2 cdot 8right) ) ( =4 p^{2} )
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