Product of numbers

Question

Pair of numbers whose product is $+7$ and whose sum is $-8$. Factorise $x^{2} - 8x + 7$.

I can factorise but it's just I can't find any products of $+7$ and that is a sum of $-8$. Any idea? Thanks guys!

Thanks.

please make it simple, lol I'm only a teengaer with raging hormones and a weird brain — , Feb 08 '13 at 18:43
@user61406: something is wrong with the signs in your question, it should be $+8x$. Please see my answer. Did you make a sign error? Regards — Amzoti, Feb 08 '13 at 18:55

score 4 · Answer 1 · answered Feb 08 '13 at 18:41

4

If you can factorize $x^2 - 8x + 7$, then you can solve a quadratic equation.

Now recall that if the roots of $x^2+bx+c=0$ are $\alpha$ and $\beta$, then $-b = \alpha + \beta$ and $c = \alpha \beta$.

(This is because $(x-\alpha)(x-\beta) = x^2 - (\alpha+\beta) x + \alpha \beta$.)

answered Feb 08 '13 at 18:41

Andreas Caranti

68,873

lol thanks, I know I was just a big confused I guess! – Feb 08 '13 at 18:54

Amzoti · Accepted Answer · 2013-02-08T18:53:11.160

2

$x y = 7$

$x + y = -8$

$y = \frac{7}{x}$, so

$x + \frac{7}{x} = -8$

Multiplying by $x$, yields:

$$x^2 +8x +7 = 0$$

$$(x+1)(x+7) = 0$$

Can you take it from there?

Regards

edited Feb 08 '13 at 18:53

answered Feb 08 '13 at 18:44

Amzoti

56,093

THANKS A LOT!!!!!!!!!!!!! LOL I only get your explanation! – Feb 08 '13 at 18:49
@user61406: you are very welcome! It helps to acknowledge people who answered questions by upvoting them and there are some nice answers. You should also accept the answer you like best. Regards – Amzoti Feb 08 '13 at 18:50
damn, well I think I got your the best. Other answers were sqaure roots and inspections etc... Smart people lol – Feb 08 '13 at 18:52
What wonderful feedback! Your answer deserves more $\uparrow$'s $\quad +1$ – amWhy May 03 '13 at 01:33
Yeah, sometimes the easiest approach is easiest! Simplicity works! :-) – Amzoti May 03 '13 at 01:34
How did you factor $x^2+8x+7$? – The Chaz 2.0 May 12 '13 at 21:11
Since $7$ is prime, its only factors are $1, 7$ and I need an $8$ as a sum, thus $(x+1)(x+7)$. Clear? Regards – Amzoti May 12 '13 at 21:13

score 0 · Answer 3 · answered Feb 08 '13 at 18:42

$(x-a)(x-b)=x^2-(a+b)x+ab$

So, you are looking for the roots of $x^2-(+2)x+(-3)$, that is, $x^2-2x-3$, which is $=x^2-2x+1-4=(x-1)^2-4$, and this is $0$ iff $(x-1)^2=4$, that is, $x-1=\pm 2$, i.e., $-1,3$ is your solution.

Similarly for the other one: $x^2-8x+7=(x-4)^2-16+7=\dots$

Or, the other way: $7+1=8$ and $7\cdot 1=7$, so the roots of $x^2-8x+7$ are $1$ and $7$.

score 0 · Answer 4 · answered Feb 08 '13 at 18:44

0

I do not understand why you are trying to factorise $x^2-8x+7$. I suggest you use viete's formulae. xy=2 and x+y=-3. Let's say you have a quadratic equation $x^2+ax+b$ Then the roots $x_1, x_2$ has the property $x_1+x_2=-a$ and $x_1.x_2=c$ So you have the quadratic equation $x^2-2x-3$ When you solve this, you get the answer.

answered Feb 08 '13 at 18:44

ciceksiz kakarot

357

solve it? lol what do you mean? like factorise that? sos – Feb 08 '13 at 18:46
yeah sorry :D I meant to solve the equation $x^2-2x-3=0$. – ciceksiz kakarot Feb 11 '13 at 18:33

iostream007 · Answer 5 · 2013-05-13T10:19:08.260

0

your questions answer are already users have given so i'm not repeating them. sometimes factorization is not easy because number are too big to manually calculate or may be fractional so try this formula as alternative. roots for equation $(ax^2+bx-c)=0$ is, $x=\dfrac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ , $\dfrac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$

edited May 13 '13 at 10:19

answered May 12 '13 at 21:06

iostream007

4,529

Product of numbers

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