what two numbers can you add to get one and multiply to get fourty-eight?

Question

it is an eleventh grade math riddle/problem but I do not know how to get the answer.

Answer 1

a +b =1 and
ab =48

b = 1 -a
a(1 -a) =48

a^2 -a +48 =0
D = 1 -4(48 ) = -191 ( no real solutions)

a1 = (1 +i sqrt(191))/(2)
a2 = (1 -i sqrt(191))/(2)

b1 = 1 -a1 = (1- i sqrt((191))/2
b2 = 1 -a2 = (1+ i sqrt((191))/2

a = (1 +i sqrt(191))/(2) and b = (1- i sqrt((191))/2

are solutions.

Answer 2

I don't believe there is an answer, but let's see:

let's let x and y be the two numbers...

if we add them, we get 1:

x + y = 1

if we multiply them, we get 48:

xy = 48

Let's use substitution:

x + y = 1 so y = 1-x

xy = 48, substitute 1-x for y to get
x(1-x) = 48
x - x^2 = 48
x^2 - x + 48 = 0

Now we can use the quadratic formula to solve for x:

x = -b +/- (b^2 - 4ac)^1/2 / 2a
x = -(-1) +/- (-1^2 - 4(48))^1/2 / 2
x = 1 +/- (1-192)^1/2 / 2

Looking inside the parenthesis, we see that we will need to take the square root of -191, so there are no real answers to this. If you want imaginary answers:

x = {1 +/- (191)^1/2 i} / 2
x = (1 +/- 13.82i) / 2

Hope this helps! (I suspect there is a typo in the original problem... perhaps a missing negative sign?)

Answer 3

Let the two numbers be a and b
then acc. to condition
a+b=1 implies b= 1-a
ab=48
put b= 1-a , we get
a( 1-a ) = 48
a - a^2 =48
a^2 -a +48 =0

a= ( 1+ sqrt( 1- 4*48))/2

which is a complex number

so no such real numbers exist for which if you add get one and multiply get fourty-eight

Answer 4

Let's call the numbers A and B:
A+B = 1
A*B = 48

Express A in terms of B:
A = 48/B

Substitute into the first equation:
48/B + B = 1

Now multiply both sides by B, to get rid of the denominator:
48 + B² = B

Group everything on one side:
B² - B + 48 = 0

Now you can solve with the quadratic equation:
a = 1, b = -1, c = 48

B = [-b +/- sqrt(b² - 4ac) ] / 2a
B = (1 +/- sqrt(1 - 4*48) ) / 2
B = (1 +/- sqrt( -191 ) ) / 2

Oooh, you are going to get into irrational numbers:
B = ½ +/- ½sqrt(191)i

So the numbers are:
1 + i√(191)
----------------
...... 2

and

1 - i√(191)
----------------
...... 2

If you add these the i√191 terms cancel out and you get:
½ + ½ = 1

If you multiply them, you get:
1² + (√191)²
----------------
....... 2²

1 + 191
-----------
..... 4

192
------
.. 4

= 48

So your answer is the following two irrational numbers:
1 + i√191 ........ 1 - i√191
-------------- and -------------
..... 2 ........................ 2

Answer 5

1+0=1x48=48

Answer 6

x+y=1
xy = 48
(1-y)y = 48
y-y^2=48
y^2-y+48 = 0
y = [1 +/- sqrt(1-4*48)]/2
y = 1/2 +/- i/2 sqrt(191)
The solution involves imaginary numbers.

Answer 7

a+b=1
ab=48
obviously, this is impossible and must be a kind of riddle.

Answer 8

impossible......