Which equation has no real roots?
\(x^2-4x+3\sqrt{2}=0\)
\(x^2+4x-3\sqrt{2}=0\)
\(x^2-4x-3\sqrt{2}=0\)
\(3x^2+4\sqrt{3}x+4=0\)
To check whether a quadratic equation has real roots or not, we use the discriminant formula:
\(D = b^2 - 4ac\).
- If \(D > 0\), there are two different real roots.
- If \(D = 0\), there are two equal real roots.
- If \(D < 0\), there are no real roots.
Option (A): \(x^2 - 4x + 3\sqrt{2} = 0\)
Here, \(a = 1, b = -4, c = 3\sqrt{2}\).
So, \(D = (-4)^2 - 4(1)(3\sqrt{2}) = 16 - 12\sqrt{2}\).
Since \(12\sqrt{2} \approx 16.97\), we get \(16 - 16.97 < 0\).
That means \(D < 0\). So no real roots.
Option (B): \(x^2 + 4x - 3\sqrt{2} = 0\)
Here, \(a = 1, b = 4, c = -3\sqrt{2}\).
So, \(D = (4)^2 - 4(1)(-3\sqrt{2}) = 16 + 12\sqrt{2}\).
This is a positive number, so there are real roots.
Option (C): \(x^2 - 4x - 3\sqrt{2} = 0\)
Here, \(a = 1, b = -4, c = -3\sqrt{2}\).
So, \(D = (-4)^2 - 4(1)(-3\sqrt{2}) = 16 + 12\sqrt{2}\).
This is also positive, so there are real roots.
Option (D): \(3x^2 + 4\sqrt{3}x + 4 = 0\)
Here, \(a = 3, b = 4\sqrt{3}, c = 4\).
So, \(D = (4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0\).
This means the roots are real and equal.
Therefore, the equation in Option (A) has no real roots.