Question

the sum of two number is 9and the sum of their squar is 41.Taking one number as x form a quadratic equation in x and solve it to find the num

Answer 1

Heya !!


Let two numbers are X and Y.


According to the question,

X + Y = 9 ---------(1)

And,


( X )² + ( Y)² = 41 ---------(2)


From equation (1) we get,


X + Y = 9


X = ( 9 - Y ) --------(3)



Putting the value of X in equation (2)



X² + Y² = 41



( 9 - Y )² + ( Y )² = 41


(9)² + ( Y )² - 2 × 9 × Y + Y² = 41

81 + Y² - 18Y + Y² = 41



2Y² - 18Y + 81 - 41 = 0



2Y² - 18Y + 40 = 0



2 ( Y² - 9Y + 20 ) = 0



Y² - 9Y + 20 = 0



Y² - 5Y- 4Y + 20 = 0



Y ( Y - 5 ) - 4 ( Y - 5 ) = 0



( Y - 5 ) ( Y - 4 ) = 0


( Y - 5 ) = 0 OR ( Y - 4 ) = 0



Y = 5 or Y = 4



Putting Y = 5 in equation (3)



X = 9 - Y = 9-5 = 4



Hence,


Numbers are 5 and 4.

Answer 2

Let the first number be x
second number =9-x
According to the question
${x}^{2} + ( x - 9)^{2} = 41 \\ {x}^{2} + {x}^{2} + 81 - 18x = 41 \\ {2x}^{2} - 18x + 81 - 41 = 0 \\ {2x}^{2} - 18x + 40 = 0 \\ {x}^{2} - 9x + 20 = 0 \\ {x}^{2} - 5x - 4x + 20 = 0\\ x(x - 5) - 4(x - 5) = 0 \\ (x - 4)(x - 5) = 0 \\ x - 4 = 0 \\ x = 4 \\ and \\ \: x - 5 = 0 \\ x = 5 \\ if \: first \: number \: 4 \: then \: second \: number = 9 - 4 = 5 \\ first \: number \: 5 \: then \: second \: number = 9 - 5 = 4 \\ required \: numbers \: are \: 4 \: and \: 5$