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Two numbers $p$ and $q$ are such that their sum is $78$ and their product is $560$. Determine the values of $p$ and $q$.

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closed as off-topic by Did, kingW3, Magdiragdag, user223391, Shailesh Nov 22 '17 at 0:16

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  • $\begingroup$ $$0=t^2-78t+560=(t-8)\cdots$$ $\endgroup$ – lab bhattacharjee Nov 20 '17 at 12:06
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To find two numbers given their sum and product is an ancient problem type dating back to antiquity. It is easy enough to work the problem into a quadratic equation in standard form where you can apply the quadratic formula (as in the other answers), but we can also attack it directly:

Since the sum is $78$, the two numbers must be the same distance above and below $\frac{78}{2}=39$, or in other words $$ p = 39 + x \\ q = 39 - x $$ for some $x$. Their product is then $$ pq = (39+x)(39-x) = 39^2 - x^2 = 1521 - x^2 $$ For this to be $560$ we must have $x^2 = 1521-560 = 961 $, and we can then find $x$ itself as $\sqrt{961} = 31 $.

So $$ p = 39+31 = 70 \\ q = 39-31 = 8 $$ (or the other way around if we had chosen the negative square root of $961$ instead).

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p+q = 78, pq = 560 By using Vieta's formulas, we call p and q roots of a certain quadratic.

x^2 - 78 x + 560 = 0.

Now, we can factor. (x-8)(x-70).

This gives us p and q as 8 and 70. Hope this helped

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$\left\{ \begin{array}{l} p+q=78\\ p\cdot q=560 \end{array} \right.$

$\left\{ \begin{array}{l} q=78-p\\ p\,(78-p)=560 \end{array} \right.$

$78p-p^2-560=0\to p^2-78p+560=0\to p_1=8;\;p_2=70$

$\left\{ \begin{array}{l} p=8\\ q=70 \end{array} \right.$ $\left\{ \begin{array}{l} p=70\\ q=8 \end{array} \right.$

Hope it helps

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From $p\cdot q=560$, we divide both sides to get $p=\frac{560}{q}$. Replace $p$ with $\frac{560}{q}$ in $p+q=78$. From $\frac{560}{q}\cdot q=78$ we multiply by $q$ and move the $78q$, giving you $q^2-78q+560=0$. Using the quadratic formula and using $p+q=78$, we can find the solution sets are: $$\left\{ \begin{array}{l} p=8, q=70\\ p=70,q=8 \end{array} \right.$$

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