Mathematics

# find the roots ? x² + 7x + 12 =0 x²-8x +12=0

#### DestinyNicolle

4 years ago

We have to remember one simple pattern $\Delta =b^2-4ac$
1.
$x^2+7x+12=0$

in this case

$b=7$
$a=1$
$c=12$

so:

$\Delta =7^2-4*1*12$
$\Delta =49-48$
$\Delta =1$
$\Delta >0$ it means that there are 2 roots

$\sqrt{\Delta} =1$

$x_{1}=\frac{-b+\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b-\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-7+1}{2}$
$x_{1}=\frac{-6}{2}$
$x_{1}=-3$

$x_{2}=\frac{-7-1}{2}$
$x_{2}=\frac{-8}{2}$
$x_{2}=-4$

Roots are $x_1=-4$ and $x_2=-3$
2.
$x^2-8x +12=0$

in this case

$a=1$
$b=-8$
$c=12$

$\Delta =(-8)^2-4*1*12$
$\Delta =64-48$
$\Delta =16$
$\Delta >0$ it means that there are 2 roots
$\sqrt{\Delta}=4$

$x_{1}=\frac{-b+\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b-\sqrt{\Delta}}{2a}$

$x_{1}=\frac{8+4}{2}$
$x_{1}=\frac{12}{2}$
$x_{1}=6$

$x_{2}=\frac{8-4}{2}$
$x_{2}=\frac{4}{2}$
$x_{2}=2$

Roots:$x_1=6$ and $x_2=4$

#### yassautumn

4 years ago

$x^2 + 7x + 12 =0\\ x^2+4x+3x+12=0\\ x(x+4)+3(x+4)=0\\ (x+3)(x+4)=0\\ x=-3 \vee x=-4\\\\ x^2-8x+12=0\\ x^2-6x-2x+12=0\\ x(x-6)-2(x-6)=0\\ (x-2)(x-6)=0\\ x=2 \vee x=6$