To determine which quadratic equation has no real solutions, we should evaluate the discriminant of each option. For a quadratic equation in the form $ax^2 + bx + c = 0$, the discriminant is given by:

$\Delta = b^2 - 4ac$

If $\Delta < 0$, the equation has no real solutions.

Let’s examine each option:

1. Option A: $x^2 + 14x - 49 = 0$

   • $a = 1$, $b = 14$, $c = -49$
   • Discriminant: $\Delta = 14^2 - 4 \cdot 1 \cdot (-49) = 196 + 196 = 392$
   • Since $\Delta > 0$, this equation has real solutions.

2. Option B: $x^2 + 14x + 49 = 0$

   • $a = 1$, $b = 14$, $c = 49$
   • Discriminant: $\Delta = 14^2 - 4 \cdot 1 \cdot 49 = 196 - 196 = 0$
   • Since $\Delta = 0$, this equation has a real (repeated) solution.

3. Option C: $5x^2 - 14x - 49 = 0$

   • $a = 5$, $b = -14$, $c = -49$
   • Discriminant: $\Delta = (-14)^2 - 4 \cdot 5 \cdot (-49) = 196 + 980 = 1176$
   • Since $\Delta > 0$, this equation has real solutions.

4. Option D: $5x^2 - 14x + 49 = 0$

   • $a = 5$, $b = -14$, $c = 49$
   • Discriminant: $\Delta = (-14)^2 - 4 \cdot 5 \cdot 49 = 196 - 980 = -784$
   • Since $\Delta < 0$, this equation has no real solutions.

Conclusion

The correct answer is Option D because it has a negative discriminant, indicating no real solutions.

Would you like further details on discriminants or have any other questions?

Related Questions:

1. How do discriminants determine the nature of solutions in quadratic equations?
2. What are complex roots, and when do they occur in quadratic equations?
3. Can a quadratic equation have exactly one real solution?
4. What are the implications of a zero discriminant in quadratic equations?
5. How do we complete the square for a quadratic equation?

Tip:

For quick determination of real solutions in quadratics, always calculate the discriminant first.