Question

Let c be a positive integer. Choose the quadratic function that ALWAYS has two irrational roots. A. f(x) = x 2– c B. f(x) = x 2– c 2 C. f(x) = cx 2– c D. f(x) = cx 2+ cx – c

Answer 1

To determine which quadratic function always has two irrational roots, we need to analyze the discriminant of each quadratic function. The discriminant \(D\) of a quadratic function in the form \(ax^2 + bx + c\) is given by \(D = b^2 - 4ac\). If \(D > 0\) and not a perfect square, the roots are irrational and distinct.

Let's evaluate each option.

**A. \(f(x) = x^2 - c\)**:
- Here, \(a = 1\), \(b = 0\), \(c = -c\).
- The discriminant is:
\[
D = 0^2 - 4(1)(-c) = 4c
\]
Since \(c\) is a positive integer, \(D = 4c\) is positive and is a perfect square (specifically \(D = (2\sqrt{c})^2\)). Therefore, this function does not always have irrational roots.

**B. \(f(x) = x^2 - c^2\)**:
- Here, \(a = 1\), \(b = 0\), \(c = -c^2\).
- The discriminant is:
\[
D = 0^2 - 4(1)(-c^2) = 4c^2
\]
Similar to option A, \(D = 4c^2\) is a perfect square. Thus, this function does not always have irrational roots.

**C. \(f(x) = cx^2 - c\)**:
- Here, \(a = c\), \(b = 0\), \(c = -c\).
- The discriminant is:
\[
D = 0^2 - 4(c)(-c) = 4c^2
\]
Again, this is a perfect square, so we can conclude that this function does not always have irrational roots.

**D. \(f(x) = cx^2 + cx - c\)**:
- Here, \(a = c\), \(b = c\), \(c = -c\).
- The discriminant is:
\[
D = c^2 - 4(c)(-c) = c^2 + 4c^2 = 5c^2
\]
Since \(5c^2\) is greater than zero and not a perfect square (as \(5\) is not a perfect square), this function will always have two distinct irrational roots.

Therefore, the quadratic function that always has two irrational roots is:

**D. \(f(x) = cx^2 + cx - c\)**.