For which interval of a will x² + 2(a − 2)x + 2a + 5 = 0 have no real roots?

Correct answer: 3 − √10 < a < 3 + √10

A quadratic equation has no real roots exactly when its discriminant is negative. For x²+2(a−2)x+2a+5=0, the coefficients are A=1, B=2(a−2), and C=2a+5. Thus D=[2(a−2)]²−4(1)(2a+5)=4(a−2)²−8a−20=4(a²−6a−1). We require D<0, so a²−6a−1<0. The corresponding equality has solutions a=3±√10. Because the coefficient of a² is positive, the expression is negative strictly between these two values. Therefore 3−√10<a<3+√10, making option A correct. At either endpoint D=0, producing equal real roots, while outside the interval D>0, producing two distinct real roots.