Question

Solve the quadratic equation using the square root method:
-108 = (x-11)^2 +13

Answer 1

`\sf x=11+11i\\x=11-11i`

Explanation:

Given quadratic equation:

`\sf -108=(x-11)^2+13`

Subtract 13 from both sides:

`\implies \sf -108-13=(x-11)^2+13-13`

`\implies \sf -121=(x-11)^2`

Switch sides:

`\implies \sf (x-11)^2=-121`

`\textsf{For $(g(x))^2=f(a)$ the solutions are $g(x)=\pm√(f(a))$}`

`\implies \sf x-11=\pm √(-121)`

Rewrite -121 as 11² · -1 :

`\implies \sf x-11=\pm √(11^2 \cdot-1)`

`\textsf{Apply radical rule} \quad √(ab)=√(a)√(b):`

`\implies \sf x-11=\pm√(11^2) √(-1)`

`\implies \sf x-11=\pm11√(-1)`

`\textsf{Apply\;imaginary\;number\;rule}:\quad √(-1)=i`

`\implies \sf x-11=\pm11i`

Add 11 to both sides:

`\implies \sf x=11\pm11i`

Therefore, the solutions are:

`\implies \sf x=11+11i`

`\implies \sf x=11-11i`