Question

The dual of (p-q)v(q-p) is​

Answer 1

Answer:

thank for giving me points

Answer 2

Answer:

9

93x

93x 2

93x 2 +ax−2=0

93x 2 +ax−2=0Since, one root is 1, then x=1

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1)

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a)

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1)

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=36

93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=36∴ b=9