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Math question

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What multiples to 192 but adds up to 16?

Guest Oct 27, 2017

#1
+302
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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: $$8^2-(8i\sqrt{2})^2=64-8^2i^2\sqrt{2}^2=64-(-128)=192$$

Mathhemathh  Oct 27, 2017
Sort:

#1
+302
+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: $$8^2-(8i\sqrt{2})^2=64-8^2i^2\sqrt{2}^2=64-(-128)=192$$

Mathhemathh  Oct 27, 2017

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