What is the value of s for equal real roots of x² − 3x + s = 0?

Correct answer: 9/4

The governing condition for equal real roots of ax²+bx+c=0 is that the discriminant must be zero: b²−4ac=0. Comparing x²−3x+s=0 with the standard form gives a=1, b=−3, and c=s. Therefore, (−3)²−4(1)(s)=0, so 9−4s=0 and s=9/4. The repeated root can independently confirm the result. Since the root is −b/(2a)=3/2, substitution gives (3/2)²−3(3/2)+9/4=9/4−9/2+9/4=0. Thus option A is correct. The values 3/4, 9, and 3 make the discriminant nonzero, so they produce two unequal roots rather than a repeated real root.