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What multiples to 192 but adds up to 16?

Guest Oct 27, 2017

Best Answer 

 #1
avatar+333 
+2

So, \(a+b=16\) (A) and \(a*b=192\). (B).

 

Multiply \(b\) to both sides of (A): \(ab+b^2=16b\).

Subtract the two equations together: \(b^2=16b-192\).

Turn it into standard form of a quadratic: \(b^2-16b+192=0\).

Unfortunately, this doesn't have any real solutions, but I'll show the complex ones anyway. \(b={16\pm16i\sqrt{2}\over 2}=8\pm8i\sqrt{2}\)

Substituting this into (A) gets us: \(a+8\pm8i\sqrt{2}=16\).

Subtract 8 from both sides: \(a\pm8i\sqrt{2}=8\).

Isolate a: \(a=8\mp8i\sqrt{2}\).

 

QUICK CHECK:

\(8\mp8i\sqrt{2}+8\pm8i\sqrt{2}=16?\)

Plus-minus and minus-plus cancel out: \(8+8=16\)\(16=16\)

 

\((8\mp8i\sqrt{2})(8\pm8i\sqrt{2})=192?\)

Product of sum and difference: \(8^2-(8i\sqrt{2})^2=64-8^2i^2\sqrt{2}^2=64-(-128)=192\) wink

Mathhemathh  Oct 27, 2017
 #1
avatar+333 
+2
Best Answer

So, \(a+b=16\) (A) and \(a*b=192\). (B).

 

Multiply \(b\) to both sides of (A): \(ab+b^2=16b\).

Subtract the two equations together: \(b^2=16b-192\).

Turn it into standard form of a quadratic: \(b^2-16b+192=0\).

Unfortunately, this doesn't have any real solutions, but I'll show the complex ones anyway. \(b={16\pm16i\sqrt{2}\over 2}=8\pm8i\sqrt{2}\)

Substituting this into (A) gets us: \(a+8\pm8i\sqrt{2}=16\).

Subtract 8 from both sides: \(a\pm8i\sqrt{2}=8\).

Isolate a: \(a=8\mp8i\sqrt{2}\).

 

QUICK CHECK:

\(8\mp8i\sqrt{2}+8\pm8i\sqrt{2}=16?\)

Plus-minus and minus-plus cancel out: \(8+8=16\)\(16=16\)

 

\((8\mp8i\sqrt{2})(8\pm8i\sqrt{2})=192?\)

Product of sum and difference: \(8^2-(8i\sqrt{2})^2=64-8^2i^2\sqrt{2}^2=64-(-128)=192\) wink

Mathhemathh  Oct 27, 2017

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