Question

Solve by Factoring

Answer 1

1)
$k^2=21+4k\\\\ k^2-4k-21=0\\\\ (k^2-2\cdot 2k+4)-4-21=0\\\\ (k-2)^2-25=0\\\\ (k-2)^2=5^2\\\\ |k-2|=5\\\\\ k-2=5 \vee k-2=-5\\\\ k=7\vee k=-3$

2)
-4k = 3k-7k

$k^2=21+4k\\\\k^2-4k-21=0\\\\k^2+3k-7k-21=0\\\\ k(k+3)-7(k+3)=0\\\\(k+3)(k-7)=0\\\\ k+3=0 \vee k-7=0\\\\ k=-3\vee k=7$

3)

$k^2=21+4k\\\\k^2-4k-21=0\\\\a=1 , b=4 , c=-21 \\\\ \Delta=b^2-4ac=4^2-4\cdot 1\cdot (-21)=16+84=100\\\\ \sqrt\Delta=10\\\\ k_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-(-4)-10}{2\cdot 1}=\frac{4-10}{2}=-3\\\\ k_2=\frac{-b+\sqrt\Delta}{2a}=\frac{-(-4)+10}{2\cdot 1}=\frac{4+10}{2}=7$
$x\in \{-3, \ 7\}$

Answer B