For which value of p will px²+6x+3=0 have equal roots?

Correct answer: p=3

A quadratic equation has equal real roots exactly when its discriminant is zero, with the coefficient of x² remaining nonzero. For px²+6x+3=0, identify a=p, b=6, and c=3. Hence D=b²−4ac=6²−4(p)(3)=36−12p. Set D=0: 36−12p=0, so 12p=36 and p=3. Since p=3 is nonzero, the coefficient of x² is nonzero and the expression is genuinely quadratic. Therefore option C is correct. Checking the distractors confirms the result: p=1 gives D=24, p=2 gives D=12, and p=4 gives D=−12. These correspond respectively to distinct real roots for the first two values and non-real roots for p=4, not equal real roots.