Ooh, a puzzle. I like these.
xy=1 AND
x+y=5 ==> y=5-x this step will keep things easier since we will need to substitute
You're going to wind up with a quadratic equation which will give you two numbers.
Substituting, we can have:
y=5-x --> x(5-x)=1 ==> 5x-x2=1 which can be rewritten as
0=x2-5x+1
Boom.
Now to solve this, we go to the quadratic equation. ax2+bx+c
b2-4ac = 25-4(1)(1)=21 I like to solve the inside of the sqrt bracket first
(-b± sqrt (21)) /2a here is the equation with the inside already completed - notice there is a plus/minus sign. there are two equations to be done, one addition, one subtraction.
Since the sqrt(21) is a crazy decimal thing, I won't resolve it into a decimal until the end.
-b+sqrt(21)/2a = 5+sqrt(21)/2
-b-sqrt(21)/2a = 5-sqrt(21)/2
sqrt(21) is approx. 4.583 so we get
5+4.583/2 = 9.583/2 = 4.792 = x
and
5-4.583/2 = .417/2 = .208 = y
Note: the numbers given as x and y are accurate but not 100% because of decimal rounding. This is a good an answer as any and no one should have cause to complain; the answer checks out. :)
1
-1
-13
-209
-9
7
1
1 + i and 1 - i, where i is the imaginary number equal to the square root of -1
Let E1 and E2 be two even numbers. Then (E1+1)(E2+1) will be the product of two odd numbers. We have E1*E2 +E2+E1+1. Now when we add or multiply even numbers, we get even numbers and we add 1, it's odd.
you multiply the one and the three and add the two
There aren't any whole numbers that satisfy these conditions. The only numbers that multiply to 23 are 23 and 1.
-1
31
The two numbers that satisfy the given conditions are 4 and 1. When you add 4 and 1, you get 5. When you multiply 4 and 1, you get 4. The two numbers are 4 and 1.
-1
-1
There are no two real numbers that do, but the complex numbers (2 + i) and (2 - i) will.If you want two numbers that multiply to negative 5 and add to positive 4 then: -1 and 5If you want two numbers that multiply to negative 5 and add to negative 4 then: 1 and -5