Question

If the equation $x^2 - 4x + a = 0$ has no real roots, then

• A. $a \leq 4$
• B. $a>4$
• C. $a<2$
• D. $a<4$

Hint:

Use the discriminant $D = b^2 - 4ac$ to determine the nature of quadratic roots: - $D>0$: Two real roots
- $D = 0$: One real root
- $D<0$: No real roots

Answer 1

The Correct Option is B

Step 1: Recall the condition for real roots.
For a quadratic equation $ax^2 + bx + c = 0$, roots are real if the discriminant $D = b^2 - 4ac \geq 0$.
Step 2: Apply to given equation.
For $x^2 - 4x + a = 0$, we have:
$a = 1$, $b = -4$, $c = a$.
Thus, $D = (-4)^2 - 4(1)(a) = 16 - 4a$.
Step 3: No real roots condition.
For no real roots, $D<0$.
\[ 16 - 4a<0 \] \[ a>4 \]
Step 4: Conclusion.
Hence, for no real roots, $a>4$.