Question

question 25 (mandatory) (1 point) determine the roots of $x^2 - 22x + 121 = 0$ to the nearest hundredth. a) -11 b) 11 c) 2 and 11 d) no real solution

Explanation:

Step1: Recognize the quadratic form

The equation \(x^2 - 22x + 121 = 0\) is a quadratic equation. Notice that it might be a perfect square trinomial. A perfect square trinomial has the form \(a^2 - 2ab + b^2=(a - b)^2\). Here, \(a = x\), \(2ab = 22x\), so \(b = 11\) (since \(2\times x\times11 = 22x\)) and \(b^2 = 121\), which matches the constant term. So we can factor the quadratic as \((x - 11)^2=0\).

Step2: Solve for x

If \((x - 11)^2 = 0\), then taking the square root of both sides, we get \(x - 11 = 0\) (since the square root of 0 is 0). Solving for \(x\), we add 11 to both sides: \(x=11\). This is a repeated root, so the root of the equation is 11 (to the nearest hundredth, it's still 11.00).

Answer:

b) 11