Which equation has two distinct real roots?
\(2x^2-3\sqrt{2}x+\dfrac{9}{4}=0\)
\(x^2+x-5=0\)
\(x^2+3x+2\sqrt{2}=0\)
\(5x^2-3x+1=0\)
To check if a quadratic equation has two distinct real roots, we use the discriminant.
The discriminant is given by: \(D = b^2 - 4ac\).
Now check each option:
(A) \(2x^2 - 3\sqrt{2}x + \tfrac{9}{4} = 0\)
Here, \(a = 2, b = -3\sqrt{2}, c = \tfrac{9}{4}\).
So, \(D = (-3\sqrt{2})^2 - 4(2)(\tfrac{9}{4}) = 18 - 18 = 0\).
Since \(D = 0\), this equation has equal roots, not distinct.
(B) \(x^2 + x - 5 = 0\)
Here, \(a = 1, b = 1, c = -5\).
So, \(D = (1)^2 - 4(1)(-5) = 1 + 20 = 21\).
Since \(D > 0\), this equation has two distinct real roots.
(C) \(x^2 + 3x + 2\sqrt{2} = 0\)
Here, \(a = 1, b = 3, c = 2\sqrt{2}\).
So, \(D = (3)^2 - 4(1)(2\sqrt{2}) = 9 - 8\sqrt{2}\).
Since \(8\sqrt{2} \approx 11.3\), we get \(D \approx -2.3 < 0\).
This means no real roots.
(D) \(5x^2 - 3x + 1 = 0\)
Here, \(a = 5, b = -3, c = 1\).
So, \(D = (-3)^2 - 4(5)(1) = 9 - 20 = -11\).
Since \(D < 0\), there are no real roots.
Conclusion: Only option (B) has \(D > 0\). So, the correct answer is Option B.