Question

47. The sum of two numbers is 30 and sum of their squaresis
458. The numbers are:
e-half
a) 14,16 b) 12,18 c) 13,17 d) 11,15​

Answer 1

its 12 and 18

see the attachment of mine

and please mark as brainliest

Answer 2

The numbers are 13 and 17 (option c).

Given,

Sum of two numbers = 30

Sum of their squares = 458

To find,

The two numbers.

Solution,

The numbers are 13 and 17.

We can easily solve this problem by following the given steps.

Now,

Let's take the two numbers to be x and y.

Then,

Sum of two numbers = 30

x + y = 30

x = (30-y) ---equation 1

Sum of their squares = 458

x² + y² = 458

Putting the value of x from equation 1,

(30-y)² + y² = 458

(30)² + (y)² - 2 ×30×y + y² = 458 [ Solving (30-y)² using (a-b)² = a²+b²-2ab]

900 + y² - 30y + y² = 458

2y² - 30y + 900 = 458

2y² - 30y + 900 - 458 = 0 (Moving 458 from R.H.S. to L.H.S.)

2y² - 30y + 442 = 0

2 (y²-15y+221) = 0 [Taking 2 common out of these three terms]

y²-15y+221 = 0

To find the value of y, we need to factorise it by splitting the middle term, 15y, into two terms such that their addition will be -15y and multiplication will be 221y².

These two terms will be -17y and -13y.

y² - 17y - 13y + 221 = 0

y(y-17) -13(y-17) = 0 [ Taking y common from the first two terms and -13 common from the last two terms.]

(y-17) (y-13) = 0

y-17 = 0 or y-13 = 0

y = 17 or y = 13

Putting these two values of y one by one in equation 1,

x = 30-y

x = 30-17 (y=17)

x = 13

x = 30-y

x = 30-13 (y = 13)

x = 17

Hence, the numbers can be (17,13) or (13,17).