Question

A quadratic equation \[ x^2-ax+b=0,\qquad a>0 \] has real roots. If the sum of the roots is less than the product of the roots, then \(b\) lies in the interval

• A. \((4,\infty)\)
• B. \((-\infty,0)\cup(4,\infty)\)
• C. \((-\infty,1)\cup(2,\infty)\)
• D. \((1,3)\)

Hint:

For the quadratic equation \[ x^2-Sx+P=0, \] \[ \boxed{\text{Sum of roots}=S,\qquad \text{Product of roots}=P.} \] Always use the discriminant condition \[ \boxed{S^2-4P\ge0} \] when the roots are real.

Answer 1

The Correct Option is A

Step 1: Find the sum and product of the roots. For the quadratic equation \[ x^2-ax+b=0, \] let the roots be \(\alpha\) and \(\beta\). Then, \[ \alpha+\beta=a, \] and \[ \alpha\beta=b. \] Given, \[ \alpha+\beta<\alpha\beta, \] which gives \[ \boxed{a<b.} \]

Step 2: Use the condition for real roots. Since the equation has real roots, \[ a^2-4b\ge0. \] Thus, \[ a^2\ge4b. \] Combining with \[ a<b, \] we obtain \[ a^2>4a. \] Since \[ a>0, \] dividing by \(a\), \[ a>4. \] Also, \[ b>a>4. \] Hence, \[ \boxed{b>4.} \] Therefore, \[ \boxed{b\in(4,\infty).} \] Hence, \[ \boxed{(A)} \] is the correct answer.