The following is the comment written for a C function./* This function computes the roots of a quadratic equation a.x^2 + b.x + c = 0. The function stores two real roots in *root1 and *root2 and returns the status of validity of roots. It handles four different kinds of cases. (i) When coefficient a is zero irrespective of discriminant (ii) When discriminant is positive (iii) When discriminant is zero (iv) When discriminant is negative. Only in case (ii) and (iii), the stored roots are valid. Otherwise 0 is stored in the roots. The function returns 0 when the roots are valid and -1 otherwise. The function also ensures root1 >= root2. int get_QuadRoots(float a, float b, float c, float *root1, float *root2); */A software test engineer is assigned the job of doing black box testing. He comes up with the following test cases, many of which are redundant. Which one of the following options provide the set of non-redundant tests using equivalence class partitioning approach from input perspective for black box testing?

Equivalence class partitioning divides the input domain into partitions of equivalent data from which test cases can be derived. A non-redundant test suite selects exactly one representative test case from each equivalence class.

From the function specification, the four distinct equivalence classes based on inputs are:

1. Class 1 (Case i): Coefficient $a = 0$ (linear equation or invalid quadratic).

2. Class 2 (Case ii): Discriminant $D = b^2 - 4ac > 0$ (two distinct real roots).

3. Class 3 (Case iii): Discriminant $D = b^2 - 4ac = 0$ (two equal real roots).

4. Class 4 (Case iv): Discriminant $D = b^2 - 4ac < 0$ (complex roots).

Now evaluate each provided test case:

• T1: $a = 0.0, b = 0.0, c = 7.0 \implies a = 0$ (Class 1)

• T2: $a = 0.0, b = 1.0, c = 3.0 \implies a = 0$ (Class 1)

• T3: $a = 1.0, b = 2.0, c = 1.0 \implies D = 2^2 - 4(1)(1) = 0$ (Class 3)

• T4: $a = 4.0, b = -12.0, c = 9.0 \implies D = (-12)^2 - 4(4)(9) = 144 - 144 = 0$ (Class 3)

• T5: $a = 1.0, b = -2.0, c = -3.0 \implies D = (-2)^2 - 4(1)(-3) = 4 + 12 = 16 > 0$ (Class 2)

• T6: $a = 1.0, b = 1.0, c = 4.0 \implies D = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0$ (Class 4)

To cover all four equivalence classes without redundancy, we must pick one test case from $\{T1, T2\}$, one from $\{T3, T4\}$, $T5$, and $T6$.

Among the options, (C) T2, T4, T5, T6 selects exactly one test case from each of the four equivalence classes.