Question

p(2)=5t²+40t+12 find p2 for this​

Answer 1

Step-by-step explanation:

Given:

The sum of two digits number and the number obtained by reversing the order of its digit = 143

Difference of digits = 3

Calculation:

Let the original number be xy = 10x + y

x – y = 3

⇒ x = 3 + y

Reversed number obtained by reversing digits of original number = yx = 10y + x

10x + y + 10y + x = 143

⇒ 10 (3 + y) + y + 10y + 3 + y = 143

⇒ 30 + 10y + y + 10y + 3 + y = 143

⇒ 22y = 143 – 33

⇒ 22y = 110

⇒ y = 110/22

⇒ y = 5

y = 5 then x = 3 + 5 = 8

∴Then the original number is 85

Answer 2

P(x) = 5t^2 + 40t +12

=D = b^2 - 4ac

D = 1600-240

D = 1360

-b ± √1360/2a

-40 - √1360/ 10

Or

-40 + √1360/10