Question

The product of two consecutive integers is 90. The equation is​

Answer 1

$\huge \underline{ \underline{ \bf \purple{Solution :}}}$

• Let's x and x + 1 be the integers

According to Question :

$\longrightarrow \: \sf( x)( \times + 1) = 90 \\ \\ \longrightarrow \: \sf {x}^{2} + x = 90 \\ \\\longrightarrow \: \underline{ \boxed{ \sf {x}^{2} + x - 90 = 0}}$

Answer 2

Step-by-step explanation:

The consecutive numbers are those numbers which come one after another. For example: 1 and 2 are consecutive, 2 and 3 are consecutive etc. and so on. Integers are the both positive and negative numbers including zero.

Let an integer be 'x'

Then, its consecutive integer is (x+1).

According to the question, x(x+1) = 90 \\ or, x²+x=90 \\ or, x² + x - 90=0 \\ or, x² + (10-9)x - 90=0 \\ or, x² + 10x - 9x -90=0 \\ or, x(x+10) -9(x+10)= 0 \\ or, (x+10)(x-9) = 0 \\ Either, x+10=0 \\ i.e. \: x=-10 \: and \: (x+1)=(-10 +1) = -9 \\ \\ OR, x-9 = 0 \\ i.e. \: x = 9 \: and \: (x+1) = (9+1) = 10

[tex] x(x+1) = 90 \\ or, x²+x=90 \\ or, x² + x - 90=0 \\ or, x² + (10-9)x - 90=0 \\ or, x² + 10x - 9x -90=0 \\ or, x(x+10) -9(x+10)= 0 \\ or, (x+10)(x-9) = 0 \\ Either, x+10=0 \\ i.e. \: x=-10 \: and \: (x+1)=(-10 +1) = -9 \\ \\ OR, x-9 = 0 \\ i.e. \: x = 9 \: and \: (x+1) = (9+1) = 10

So, the required integers are -9 and -10 or 10 and 9.

Thank You