Question

The interval that gives all possible values of the expression $\sqrt{x^2-4x+20}$ is

• A. [4,)
• B. [16,)
• C. (,-4)
• D. (16,)

Answer 1

The correct option is A

[4,)

$x^2$ - 4x + 20 = $x^2$ - 2(2)(x) + 4 - 4 + 16

= $(x-2{)}^2$ + 16

We know that $(x-2{)}^2$ ≥ 0

$(x-2{)}^2$ + 16 ≥ 16

That is , the minimum value of $x^2$ - 4x + 20 is 16 when x = 2 and it extends to $\infty$.

Hence, 16 ≤ $(x-2{)}^2$ + 16 < $\infty$

(or)

4 ≤ $\sqrt{(x-2{)}^2+16}$ < $\infty$

So, range of $\sqrt{x^2-4x+20}$ is [4,$\infty$)