Question

A Two digit no. is such that the product of the digits is 12. when 36 is added to the ,the digits interchange their places.Formulate the quadratic equations whose roots are digits of number.

Answer 1

let number = 10x + y

according to question,

xy = 12 --------(1)

again,

10x + y + 36 = 10y + x
9x - 9y = 36
x - y = 4
x = 4 + y

put this in equation (1)

y( y + 4) = 12

y² + 4y -12= 0
hence, quadratic equation ,
is y² + 4y -12 = 0

solve this y = 6, but y ≠ -2
then , x = 2

so,number= 10×2 + 6 =26

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the unit’s place digit be x

And the ten’s place digit by y.

Number = 10y + x

Interchanged Number = 10x + y

According to the Question,

⇒ xy = 12

⇒ x = 12/y ..... (i)

⇒ 10y + x + 36 = 10x + y

⇒ 10y + x - 10x - y  =  - 36

⇒ 9y - 9x = - 36

⇒ 9(y - x) = - 36

⇒ y - x = - 36/9

⇒ y - x = - 4 ....(ii)

Putting Eq (i) value in Eq (ii), we get

⇒ y - 12/y = - 4

⇒ y² - 12/y = - 4

⇒ y² + 4y - 12 = 0

⇒ y² + 6y - 2y – 12 = 0

⇒ y(y + 6) - 2(y + 6) = 0

⇒ (y + 6)(y - 2) = 0

⇒ y =  - 6, 2 (Neglecting negative sign's once)

⇒ y = 2

Putting y's value in Eq (i), we get

⇒ x = 12/y

⇒ x = 12/2

⇒ x = 6

Number = 10y + x = 10 × 2 + 6 = 20 + 6 = 26