A quadratic equation typically takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are real numbers and \( a \neq 0 \). The solutions to this equation can be understood through the discriminant, given by \( D = b^2 - 4ac \).
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Can a quadratic equation have two real roots?
Yes, a quadratic equation can have two real roots. This occurs when the discriminant is positive, \( D > 0 \).
Example: Consider the quadratic equation \( x^2 - 5x + 6 = 0 \).
The discriminant is calculated as follows: \[ D = (-5)^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \] Since \( D > 0 \), this quadratic has two distinct real roots. To find the roots, you can factor or use the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{5 \pm 1}{2} \] The roots are \( x = 3 \) and \( x = 2 \).
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Can a quadratic equation have two non-real roots?
Yes, a quadratic equation can have two non-real roots, which occurs when the discriminant is negative, \( D < 0 \).
Example: Consider the quadratic equation \( x^2 + 2x + 5 = 0 \).
The discriminant is calculated as follows: \[ D = (2)^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 \] Since \( D < 0 \), this quadratic has two complex (non-real) roots. To find the roots: \[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i. \] The roots are \( -1 + 2i \) and \( -1 - 2i \).
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Can a quadratic equation have one real solution and one non-real solution?
No, a quadratic equation cannot have one real solution and one non-real solution. The solutions to a quadratic equation must either be two distinct real roots, one real root (which is also a double root), or two non-real roots. This is a consequence of the nature of quadratic equations:
- If \( D > 0 \), there are two distinct real roots.
- If \( D = 0 \), there is one real root (a double root).
- If \( D < 0 \), there are two complex conjugate roots, which are non-real.
Thus, it is not possible to have a combination of one real and one non-real root for a quadratic equation.
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