Factors of $25$ are; $1, 5, 25$
$$5 \times 5 = 25$$ $$5 + 5 = 10$$
$$-5 \times 5 = -25$$ $$-5 + 5 = 0$$
How can I solve this?
Thanks
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$\begingroup$ Hint: this is equivalent to the system of equations $x+y = -10$, $xy = -25$. $\endgroup$– user90667Jun 7, 2015 at 19:47
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$\begingroup$ They are not whole numbers. $\endgroup$– Akiva WeinbergerJun 7, 2015 at 19:54
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4 Answers
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Solve: $$\begin{align} xy &=-25 \\ x+y &= -10 \end{align} $$
In particular, substituting $y=-10-x = -(10+x)$ in the first equation, you get $$x^2+10x -25 = 0 $$ which is an equation you should know how to handle. The two solutions are $x=-5+\sqrt{50} = 5(\sqrt{2}-1)$ and $x=-5(\sqrt{2}+1)$, from which you get respectively $y=-5(\sqrt{2}+1)$ and $y=5(\sqrt{2}-1)$.
answered Jun 7, 2015 at 19:49
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$$xy = -25$$ and $$x + y = -10$$ and so from the second equation we have $$y = -10 -x$$ and we can substiute that value of $y$ in the very fisrt eqyatuoon to get $$x(-10-x) = -25$$ which is the same as $$-10x-x^2 =-25$$
and we multiply both sides by $-1$ to get $$x^2 + 10x = 25$$ and so $$x^2 + 10x -25 = 0$$ and so solve this and get back the value of $x$ to solve for $y$ and you are done !!
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Let the numbers be $a$ and $b$. Then $ab=-25$ and $a+b=-10$, from which it follows that they are the roots of the quadratic equation $$x^2+10x-25=0$$
answered Jun 7, 2015 at 19:49
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Well one cannot solve this if only integers may be used.
The answer is: $$(a,b) = (5(\sqrt{2} - 1), -5(1+\sqrt{2}))$$
We can see this as we have $$a*b = -25,\qquad a+b =-10.$$ From this it follows that $a = -10-b$, thus $$-10b -b^2 = -25, \Rightarrow b^2+10b-25 = 0.$$ This can be solved quite easily, as this gives $$b = \frac{-10\pm \sqrt{10^2-4*-25}}{2} = -5 \pm 5\sqrt{2}.$$
