Question

What two numbers add to 18 and multiply to 82

Answer 1

`x^2 +18x + 82\\\\ (-b \pm √( b^2 -4( a)(c)) )/(2(a))\\\\ (-18\pm √(18^2 -4(1)(82)) )/(2(1))\\\\ (-18\pm √(-4 ) )/(2) = (-18\pm √(-4*1 ) )/(2) \\\\ = (-18\pm 2i√(1 ) )/(2) \\\\ = (-9\pm 1i√(1 ) )/(1) \\\\ = -9\pm 1i√(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

Answer 2

The answers are complex numbers...

Here's why...

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p+q=18

pq=82

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Therefore:

p=18-q

p=82/q

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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

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Therefore:

q=9-i

q=9+i

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ANSWERS:

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

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

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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

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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

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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