Gustavo Lam

2022-04-08

I'm wrestling with this quadratic and trying to figure out how to factor it:
${a}^{2}-9a+14=0$

Varikoseocaz

Step 1
So here is a trick. The last term, 14, put into your calculator as $\frac{14}{x}$. No see what what two numbers will give you 14.
You should see that the two numbers that you should pick have the property that if you mutiple them it will give you 14 and if you add them they should give you your middle number -9. So,
$-2×-7=14$
and$-2±7=-9$
meaning:
${a}^{2}-9a+14=0$

Frain4i62

Step 1
Here is a trick that is about 2000 years old. It's actually kind of advanced, but also kind of interesting
${x}^{2}-9x+14=\left(x-r\right)\left(x-s\right)={x}^{2}-\left(r+s\right)x+rs$
So you are looking for numbers r and s such that
$r+s=9$ and $rs=14$
Let $r=\frac{9}{2}-t$ and $s=\frac{9}{2}+t$
I know this doesn't make any sense, but, stay with me here.
Clearly $r+s=9$ So we have already solved that part.
Next we work on $rs=14$
$rs=14$
$\left(\frac{9}{2}-t\right)\left(\frac{9}{2}+t\right)=14$
$\frac{81}{4}-{t}^{2}=14$
$81-4{t}^{2}=56$
$4{t}^{2}=25$
$2t=±5$$t=±\frac{5}{2}$
So, $\left\{r,s\right\}=\left\{\frac{9}{2}+\frac{5}{2},\frac{9}{2}-\frac{5}{2}\right\}=\left\{7,2\right\}$

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