Question

( 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} )

# 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.

Solution