Question

Solve by Factoring


`k^(2)=21+4k`

A) {7} B) {-3, 7}

C) {-7,0} D) {8,2}

(plz show me step by step how you solved it.)

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=(-b-\sqrt\Delta)/(2a)=(-(-4)-10)/(2\cdot 1)=(4-10)/(2)=-3\\\\ k_2=(-b+\sqrt\Delta)/(2a)=(-(-4)+10)/(2\cdot 1)=(4+10)/(2)=7`

`x\in \{-3, \ 7\}`

Answer B