Question

two digit number is such that the product of their digits is 12 and 36 is added to number of digits interchange their places. Formulate quadratic equation whose roots are digits of number, also find the number​

Answer 1

Answer:

=26

Step-by-step explanation:

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✰

☛GiveN

• There is an two digit number
• Product of digits = 12
• Adding 36, reverses it

☛To FinD

• The two digit number.

✰Solution✰

❋Let the digits of the two digit number be x, y

♦️According to question :

$\large{ \sf{ \rightarrow \: \: \: xy = 12........(1)}} \\ \\ \large{ \rightarrow \: \sf{ \: \: y = \frac{12}{x} }}$

So, unit digit is 12/x

♦️Then our number is 10x + 12/x

Secondly, we have

If we add 36 to the number, it will be reversed

☛Then,

$\large{ \sf{ \rightarrow \: 10x + \frac{12}{x} = \: 10 \times \frac{12}{x} + x}} \\ \\ \large{ \sf{ \rightarrow \: 10x + \frac{12}{x} + 36 = \frac{120}{x} + x}} \\ \\ \large{ \sf{ \rightarrow \: 10x - x + \frac{12}{x} - \frac{120}{x} + 36 = 0}} \\ \\ \large{ \sf{ \rightarrow \: 9x - \frac{108}{x} + 36 = 0}} \\ \\ \large{ \sf{ \rightarrow \: 9 {x}^{2} - 108 + 36x = 0}} \\ \sf{ \underline{ \underline{ \dag{ \green{ \: \: taking \: 9 \: common}}}}} \\ \\ \large{ \sf{ \rightarrow \: \boxed{ \red{ {x}^{2} + 4x - 12 = 0}}}}$

Hence, Required Quadratic equation = x^2+4x-12=0

☛Solving the quadratic equation,

$\large{ \sf{ \rightarrow \: {x}^{2} +4x - 12 = 0}} \\ \\ \large{ \sf{ \rightarrow \: {x}^{2} + 6x - 2x - 12 = 0}} \\ \sf{ \underline{ \underline{ \dag{ \green{ \: \: by \: middle \: term \: factorisation }}}}} \\ \\ \large{ \sf{ \rightarrow \: x(x + 6) - 2(x + 6) = 0}} \\ \\ \large{ \sf{ \rightarrow \: (x + 6)(x - 2) = 0}} \\ \\ \large{ \sf{ \rightarrow \: x = - 6 \: or \: 2}} \\ \\ \large{ \sf{ \star{ a \: digit \: can \: never \: be \: negative}}} \\ \\ \large{ \sf{ \rightarrow{ \boxed{ \red{x = 2 \: \: (ten \: digit)}}}}} \: \: \\ \\ \large{ \sf{ \rightarrow \: unit \: digit = \frac{12}{x} }} \\ \\ \large{ \sf{ \rightarrow \: \boxed{ \red{ \frac{12}{2} = 6}}}}$

☛Required number = 26

✰So final Answer✰

$\large{ \boxed{ \bf{quad. \: equation = {x}^{2} + 4x - 12 = 0}}} \\ \\ \large{ \boxed{ \bf{two \: digit \: number = 26}}}$