Question

question 25 (mandatory) (1 point) ✔ saved 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. Recall that \((a - b)^2=a^2 - 2ab + b^2\).

Step2: Factor the quadratic

For \(x^2-22x + 121\), we have \(a = x\) and \(2ab=22x\), so \(b = 11\) (since \(2\times x\times11 = 22x\)) and \(b^2=121\). So the quadratic factors as \((x - 11)^2=0\).

Step3: Solve for x

Taking the square root of both sides, we get \(x - 11=0\), so \(x = 11\). This is a repeated root, and to the nearest hundredth, it is still 11.

Answer:

b) 11