Question

the difference between squares of two numbers is 120. the square of smaller number is twice the greater number. find the numbers​

Answer 1

Let the smaller number and larger number be x and y.

x^2 = 2y …… (1)

And,

y^2 - x^2 =120  ……… (2)

Keep (1) in (2), we get

y^2 - 2y = 120

y^2 - 2y - 120 = 0

y^2 + 10y - 12b - 120 = 0

y(y + 10) - 12(y - 10) = 0

y = 12,-10(Neglect negative value)

y = 12

Then put in (1), we get

x^2 = 2y

x = √24

Therefore, the values are 12 and √24.

Answer 2

Let ,

The two numbers are x and y

By the given condition ,

$\star$The difference between squares of two numbers

(x)² - (y)² = 120 ------- eq (i)

$\star$The square of smaller number is twice the greater number

(y)² = 2x ------- eq (ii)

Put the value of (x)² = 2y in eq (i)

$\implies \sf x - { (2x) }^{2} = 120 \\ \\ \sf \implies</p><p>x - 2 {x}^{2} + 1 = 120 + 1 \\ \\ \sf \implies</p><p> {(x - 1)}^{2} = 121 \\ \\ \sf \implies</p><p>{(x - 1)}^{2} = {(11)}^{2} \\ \\\sf \implies</p><p>x - 1 = ± \: 11 \\ \\ \sf \implies </p><p>x= \: 1 ± \: 11 </p><p>$

So , we get

x = 1 + 11 & x = 1 - 11 i.e x = 12 & x = -10

Put the value of x = 12 in eq (ii), we get

$\sf \implies {(y)}^{2} = 2 × 12 \\ \\ \sf \implies</p><p>{(y)}^{2} = 24 \\ \\ \sf \implies</p><p>{(y)}^{2} = {(2)}^{2} × 6</p><p> \\ \\ \sf \implies </p><p>y = ± \: 2 \sqrt{6}$

Again , put the value of p = -10 in equation (ii)

$\sf \implies {(y)}^{2}= 2 × (-10) \\ \\ \sf \implies </p><p> {(y)}^{2} = -20 \\ \\ \sf \implies y = \sqrt{ - 20}$

Hence , the required numbers are x = 12 & y = ± 2√6 or x = -10 and √-20