Question

the difference of the square of two positive integer is 180 the square of the smaller number is 8 times the larger find the number​

Answer 1

Step-by-step explanation:

Let the larger number = x

Then the square of the smaller number = 8 times the larger number = 8x

and the square of the larger numbe r = x

According to the question,

x - 8x = 180

=> x - 8x - 180 = 0

=> x - 18x + 10x - 180 = 0

=> x(x - 18) + 10(x - 18) = 0

=> (x - 18) (x + 10) = 0

=> x - 18 = 0 or x + 10 = 0

=> x = 18 or x = -10

Thus, the larger number = 18 or -10

Then, the square of the smaller number = 8(18) or 8(-10)

= 144 or -80

The square of a number can't be negative, so, the square of smaller number = 144

Hence, the smaller number = sqrt(144) = 12

The numbers are 12 and 18

Answer 2

Given:

• Difference of the squares of two positive integers is 180 the square of the smaller number is 8 times the larger.

To Find:

• the numbers?

$\large{ \orange{\underline{\underline{ \sf \maltese{\:understanding \: the \: question : }}}}}$

➢ here there are 2 numbers given where the square of the smaller number is 8 times the larger and the difference of their squares is 180.

Solution:

➢ Now let the smaller number x and larger number he y

Equation 1 :

$\longmapsto \sf \: {y}^{2} - {x}^{2} = 180$

As given that the square of larger number is 8 times the smaller number

➜ y² = 8x

Now let's frame an equation accordingly:

$\longrightarrow \sf \: {x}^{2} - 8x - 0 = 180 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf {x}^{2} -1 8x + 10x - 180 = 0 \\ \\ \\ \longrightarrow \sf \: x(x - 18) \:\: 10(x - 18) \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf (x - 18)(x + 10) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

${ : \implies} \tt \: x = 18 \: \: \\ \\ { : \implies} \tt \: x = - 10$

As it is said that the numbers are positive so let's discard "-10"

$\longrightarrow \sf {y}^{2} = 8x \: \: \: \: \: \: \\ \\\longrightarrow \sf {y}^{2} = 8(18) \\ \\ \longrightarrow \sf {y}^{2} = 144 \: \: \: \: \\ \\ \longrightarrow \sf \: y = \sqrt{144} \\ \\ \longrightarrow \bf \orange{y = 12 \bigstar}$

$\therefore{ \underline{ \rm{ \: the \: number \: is \: 12 \dag}}}$