Question

find the value of '(4^(0)+4^(-1))××2x​

Answer 1

Answer:

See the attachment for answer

Answer 2

Step-by-step explanation:

Since it is given that the quadratic equation x2−(p+2)x+4=0 has equal roots, therefore, b2−4ac=0 where a=1,b=(p+2) and c=4:b2−4ac=0

⇒(p+2)2−(4×1×4)=0

⇒[p2+22+(2×p×2)]−16=0(∵(a+b)2=a2+b2+2ab)

⇒p2+4+4p−16=0

⇒p2+4p−12=0

⇒p2+6p−2p−12=0

⇒p(p+6)−2(p+6)=0

⇒(p−2)(p+6)=0

⇒p−2=0,p+6=0

⇒p=2,p=−6

⇒p=2,p=−6Hence, p=2,−6.

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