two numbers whose sum and multiplication is 1?

Answer 1

vijay, if u r in 11th u will be able 2 understand this,
let no be x and y
xy=1 therefore x=1/y
x+y=1 therefore x+1/x=1
x(*2)-x+1=0
therefore solving quadratic eqn
x=(1+-root(3)i)/2
y=1/x solve
if its the best answer plz give me best answer points

Answer 2

a million enable us to assert y is the style of oranges you may desire to purchase and x is the quantity of funds you on account that, oranges are 25 cents then y cases 25 would be x so 25y=x and consequently y=x/25 2 Y is your salary consistent with hour and x is hours so y cases x would be how lots your make consequently xy=one thousand and y=one thousand/x 3 enable us to represent the two numbers as x and y so x+y=40 4 now y is 4 much less then 4x so y =4x-4 so 4x-4 +x=40 4 so 5x-4=40 4 5x=40 8 so x=9.6 so one selection is 9.6 the different might desire to be 34.4, multiply 9.6 by using 4 and subtract 4 and you in addition to mght get 34.4 so the numbers are 9.6 and 34.4 4 24 and 12 because of the fact 5 years in the past the sum of the an prolonged time grew to become into 26 on account that the two certainly one of them have elderly 5 years the hot sum is 36, divide 36 by using 3 and you get 12 multiply twelve by using 2 and you get 24 and easily to double verify you probably did it properly suited you upload them the two to be sure the sum remains the comparable 5 ok so we take the equation for the edge of a rectangle 2w+2l=p now the size is 7 greater then the width or w+7=l we now replace the variables interior the equation and resolve 2w+2(w+7)=seventy 8 distribute the two 2w+2w+14=seventy 8 integrate like words 4w+14=seventy 8 subtract the 14 4w=sixty 4 divide by using 4 w=sixteen and sixteen+7 is 23 so the width is sixteen and the size is 23

Answer 3

Let x = first number
Let y = second number

x + y = 1
xy = 1

If xy = 1, then
x = 1/y

1/y + y = 1
1/y + y/1 = 1
1 + y^2 = 1
y^2 = 0

No number squared can equal zero, so y is nonexistant. Therefore there are no two numbers whose sum and multiplication is 1.

Answer 4

1 & 1

Answer 5

Let the numbers be x and y.

x+y = 1 => y = 1-x
xy = 1

Substitute y=1-x into xy=1,
x(1-x) = 1
x - x^2 = 1
0 = x^2 - x + 1

which cannot be solved as the determinant is less than zero;
i.e. b^2 - 4ac = (-1)^2 - (4)(1)(1) = 1 - 4 = -3

We have thus proven that 2 real numbers satisfying the above requirements cannot exist.

Answer 6

There are no ordinary (real) numbers that solve your problem, but using complex numbers, you come to the solution
x = 1/2 + ai
y = 1/2 - ai
where is the imaginary number sqrt(-1)
and a is sqrt(3/4)
(sqrt means squareroot).

You can verify this:
x + y = 1/2+1/2+ai-ai = 1
xy = 1/2*1/2-ai*ai
=1/4-(-1*3/4)=1/4+3/4=1

You only have to accept the existence of the the imaginary number sqrt(-1)

Look up "imaginary numbers" or "complex numbers"
in Wikepedia.

Answer 7

(1) x + y = 1
(2) x.y = 1

x = 1 - y
(2) => (1-y)y = 1
<=> y - y^2 - 1 = 0
<=> y^2 - y + 1 = 0
=> cant find y

not such numbers

Answer 8

X+Y +1
X*Y =1
so X+ 1/y
x²-x+1 =0
2 solutions
x1 = (1+sqr(3)i)/2 y1 =(1-sqr(3)i).2
x2 = (1-sqr(3)i)/2 y2= (1+sqr(3)i)/2

Answer 9

SUM
0+1
MULTIPLICATION
1*1

Answer 10

there r no sch numbers..!