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Effectus [21]
3 years ago
15

What two numbers add to 18 and multiply to 82

Mathematics
2 answers:
Irina-Kira [14]3 years ago
6 0
x^2 +18x + 82\\\\  \frac{-b \pm  \sqrt{ b^2 -4( a)(c)} }{2(a)}\\\\  \frac{-18\pm  \sqrt{18^2 -4(1)(82)} }{2(1)}\\\\   \frac{-18\pm  \sqrt{-4 } }{2} = \frac{-18\pm  \sqrt{-4\times1 } }{2} \\\\ = \frac{-18\pm  2i\sqrt{1 } }{2} \\\\ = \frac{-9\pm  1i\sqrt{1 } }{1} \\\\ = -9\pm  1i\sqrt{1 }\\\\\boxed{\bf{=-9 \pm i}}

The two numbers that add to 18 and multiply to 82 are 9 + i and 9 - i.

(9 + i)(9 - i) = 9^2 - i^2 = 81 - (-1) = 81 + 1 = 82
(9 + i) + (9 - i) = 9 + i + 9 - i = 9 + 9 + i - i = 18
Feliz [49]3 years ago
3 0
The answers are complex numbers...

Here's why...

---------

p+q=18

pq=82

---------

Therefore:

p=18-q

p=82/q

------------

Therefore:

18-q=82/q

q(18-q)=q(82/q)

18q-q²=82

(-1)(18q-q²)=82(-1)

q²-18q=-82

(q-9)²-9²=-82

(q-9)²-81=-82

(q-9)²=-82+81

(q-9)²=-1

q-9=-√(-1)=-i

q-9=√(-1)=i

-----------

Therefore:

q=9-i

q=9+i

-------------

ANSWERS:

When q=9-i, p=9+i

When q=9+i, p=9-i

-----------

Proof:

p+(9-i)=18

p+9-i=18

p=18-9+i

p=9+i

....

p+(9+i)=18

p+9+i=18

p=18-9-i

p=9-i

--------------

More proofs:

p=82/(9+i)

p=(82/(9+i))*((9-i)/(9-i))

p=(82(9-i))/(81-9i+9i-i²)

p=(82(9-i))/(81-(-1))

p=(82(9-i))/82

p=9-i

---------------

p=82/(9-i)

p=(82/(9-i))*((9+i)/(9+i))

p=(82(9+i))/(81+9i-9i-i²)

p=(82(9+i))/(81-(-1))

p=(82(9+i))/82

p=9+i
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