Which of the following equations has no real root ?

Question

Accepted Answer

To determine which of the given equations has no real roots, we need to analyze the discriminant of each quadratic equation. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $D = b^2 - 4ac$. A quadratic equation has no real roots if its discriminant is less than zero ($D < 0$). Let's evaluate the discriminant for each option: 1. **Option 1: $x^2 - 4x + 3\sqrt{2} = 0$** - Here, $a = 1$, $b = -4$, and $c = 3\sqrt{2}$. - The discriminant $D = (-4)^2 - 4 \times 1 \times 3\sqrt{2} = 16 - 12\sqrt{2}$. - Since $12\sqrt{2} \approx 16.97$, we have $16 - 12\sqrt{2} < 0$. - Therefore, the discriminant is negative, indicating that this equation has no real roots. 2. **Option 2: $x^2 + 4x - 3\sqrt{2} = 0$** - Here, $a = 1$, $b = 4$, and $c = -3\sqrt{2}$. - The discriminant $D = 4^2 - 4 \times 1 \times (-3\sqrt{2}) = 16 + 12\sqrt{2}$. - Since $12\sqrt{2} \approx 16.97$, we have $16 + 12\sqrt{2} > 0$. - The discriminant is positive, indicating that this equation has real roots. 3. **Option 3: $x^2 - 4x - 3\sqrt{2} = 0$** - Here, $a = 1$, $b = -4$, and $c = -3\sqrt{2}$. - The discriminant $D = (-4)^2 - 4 \times 1 \times (-3\sqrt{2}) = 16 + 12\sqrt{2}$. - Since $12\sqrt{2} \approx 16.97$, we have $16 + 12\sqrt{2} > 0$. - The discriminant is positive, indicating that this equation has real roots. 4. **Option 4: $3x^2 + 4\sqrt{3}x + 4 = 0$** - Here, $a = 3$, $b = 4\sqrt{3}$, and $c = 4$. - The discriminant $D = (4\sqrt{3})^2 - 4 \times 3 \times 4 = 48 - 48 = 0$. - The discriminant is zero, indicating that this equation has exactly one real root (a repeated root). In conclusion, the only equation with a negative discriminant, and thus no real roots, is Option 1: $x^2 - 4x + 3\sqrt{2} = 0$. This supports the given correct answer.