Question

Divide 12 into two parts such that the sum of their squares
is 80

Answer 1

Answer:

So the numbers are 8 and 4

Step-by-step explanation:

We Have :-

Sum of two numbers = 12

Sum those numbers square = 80

To Find :-

Those numbers

Method Used :-

Middle term splitting method

Formula used :-

( a - b )² = a² + b² - 2ab

Solution :-

Let the numbers be x and y

Acc to ques :-

x + y = 12

y = 12 - x ------- ( i )

x² + y² = 80

x² + ( 12 - x )² = 80   [ From ( i ) ]

x² + 144 + x² - 24x = 80

2x² - 24x + 144 = 80

2x² - 24x + 144 - 80 = 0

2x² - 24x + 64 = 0   [ Dividing by 2 ]

x² - 12x + 32 = 0

x² - 8x - 4x + 32 = 0

x ( x - 8 ) - 4 ( x - 8 ) = 0

( x - 8 ) ( x - 4 ) = 0

x = 8 , 4

When

x = 8

y = 4

When

x = 4

y = 8

So the numbers are 8 and 4

Answer 2

Answer:

4 and 8

Step-by-step explanation:

$x + y = 12$

$x = 12-y$

$x^{2} +y^{2} = 80$

$(12-y)^2 + y^2 = 80$

$144 + y^2 -24y +y^2 = 80$

$2y^2 - 24y+144 =80$

$2y^2 - 24y +144 - 80 =0$

$2y^2 - 24y +64 = 0$

$y^2 -12y+32 = 0$

$y^2-8y-4y+32=0$

$y(y-8)-4(y-8)=0$

$(y-4)(y-8)=0$

(y-4)=0                                    or                                     y-8 = 0

y=4                                         or                                      y=8

x = 12 -y

=> x = 12 -4                            or                                      x = 12-8

=> x = 8                                  or                                        x = 4

Hence the numbers are 4 and 8.