Question

Sum of two numbers is 40 and their reciprocals is 2/5. represent this in the quadratic equation​. Please it's urgent

Answer 1

Correct QuestioN :

Sum of two number is 10 and Sum of their reciprocal is 5 / 12. Find the number.

SolutioN :

Let,

• The number be x and y

Case 1.

• Sum of two numbers is 10.

$\tt \dagger \: \: \: \: \: x + y = 10 \: \: \: \: \: - (1)$

Case 2.

• Sum of their reciprocals is 5 / 12.

$\tt \dagger \: \: \: \: \: \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{5}{12} \: \: \: \: \: - (2)$

Taking Equation ( 2 )

$\tt : \implies \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{5}{12}$

$\tt : \implies \dfrac{y + x}{xy} = \dfrac{5}{12}$

$\tt : \implies \dfrac{x + y}{xy} = \dfrac{5}{12}$

$\tt : \implies \dfrac{10}{xy} = \dfrac{5}{12}$

$\tt : \implies \dfrac{ \cancel{10}}{xy} = \dfrac{ \cancel5}{12}$

$\tt : \implies \dfrac{ 2}{xy} = \dfrac{ 1}{12}$

$\tt : \implies xy = 12 \times 2$

$\tt : \implies xy = 24.$

Now,

• ( x - y )² = ( x + y )² - 4xy.

$\tt \dashrightarrow{\bigg(x - y \bigg)}^{2} = {(x + y)}^{2} - 4xy.$

$\tt \dashrightarrow{\bigg(x - y \bigg)}^{2} = {10}^{2} - 4(24).$

$\tt \dashrightarrow{\bigg(x - y \bigg)}^{2} = 100 - 96$

$\tt \dashrightarrow{\bigg(x - y \bigg)}^{2} = 4$

$\tt \dashrightarrow x - y = \sqrt{4}$

$\tt \dashrightarrow x - y = \pm2.$

By Elimination Method :

x + y = 10.

x - y = 2.

_________

2x = 12.

x = 6.

Now, Putting the value of x in Equation ( 1 )

→ x + y = 10.

→ 6 + y = 10.

→ y = 10 - 6.

→ y = 4.

Therefore, the number is 6 and 4.

$\rule{200}3$

QuestioN :

• Sum of two number is 40.

$\tt \dagger \: \: \: \: \: x + y = 40 \: \: \: \: \: - (1)$

• their reciprocals is 2/5.

$\tt \dagger \: \: \: \: \: xy = \dfrac{2}{5} \: \: \: \: \: - (2)$

Taking Equation ( 2 )

$\tt : \implies xy = \dfrac{2}{5}$

$\tt : \implies x = \dfrac{2}{5y} \: \: \: \: \: - (3)$

Now, Putting the value of x in Equation ( 1 )

$\tt : \implies x + y = 40.$

$\tt : \implies \dfrac{2}{5y} + y = 40.$

$\tt : \implies \dfrac{2 + {5y}^{2} }{5y} = 40.$

$\tt : \implies 2 + {5y}^{2} = 40 \times 5y.$

$\tt : \implies 2 + {5y}^{2} = 200y.$

$\tt : \implies {5y}^{2} - 200y + 2 = 0.$

Therefore, the become as 5y² - 200y +2 = 0.