Question

The product of two consecutive positive integers is 306 . find the integers.

Class 10th

Quadratic equation.

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

x = 17 

x = 18

$\bf \pink{step \: by \: step \: explanation}$

Let x and x + 1 be the two consecutive positive integers .

Then, x(x + 1) = 306

→ x² + x - 306 = 0.

→ x² + 18x - 17x - 306 = 0.

→ x ( x + 18 ) - 17 ( x + 18 ) = 0

→ ( x - 17 ) ( x + 18 ) = 0.

→ x - 17 = 0 

→ x = 17

and 

→ x + 18 = 0 

→ x = -18 

Rejecting x = -18 , as the given numbers are positive , we get x = 17 



•°• The two integers are 17 and 18. 


We Can also find it by another method

which is Quadratic formula .

First ,

→ D = b²-4ac

and here

→ a = 1.

→ b = 1

and

→ c = -306

now

→ D = (1)² - 4×1 ×-306

→ D = 1 - 4 × -306

→ D = 1+1224

→ D = 1225.


Now

→ x = -b ±√D/2a

TAKING D POSITIVE



→ x = -1 + √1225 /2×1

→ x = - 1 + 35/2

→ X = 34/2

→ X = 17

AND TAKING D NEGATIVE.

→ x = -b - √D/2a

→ x = -1 - √1225/2×1

→ x = -1 -35/2

→ x = -36/2

→ x = -18 (it is rejected as the value is in negative )

So

integers are

17 and 17+1=18

Answer 2

$\sf{\large{\underline {QUADRATIC\:EQUATIONS}}}$

Let the two positive consecutive integers be ( x ) and ( x + 1 ).

According to the question,

Their Product = 306

x ×( x + 1 ) = 306

${x}^{2}$ + x - 306 = 0

${x}^{2}$ + 18x - 17x - 306 = 0

x ( x + 18 ) - 17 ( x + 18 ) = 0

( x + 18 ) ( x - 17 ) = 0

( x + 18 ) = 0 , ( x - 17 ) = 0

x = - 18, 17

So, admissible value of x is 17.

First Positive integer = x = 17.

Second Positive integer = ( x + 1 ) = 17 + 1 = 18.

$\sf{\underline {\large{The \:integers\: are \: 17 \: and \: 18}}}$