Answer :
Certainly! Let's consider the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
### Step-by-Step Solution:
1. Degree of the Polynomial:
A quadratic equation is a polynomial of degree 2. This means that the highest power of the variable [tex]\( x \)[/tex] is 2.
2. Fundamental Theorem of Algebra:
The Fundamental Theorem of Algebra states that every non-zero polynomial equation of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] solutions in the set of complex numbers, counting multiplicities. Since the degree of our quadratic equation is 2, it must have exactly 2 solutions in the complex number set.
3. Nature of Complex Zeros:
Complex roots (or zeros) of polynomials with real coefficients always occur in conjugate pairs. This means if [tex]\( a + bi \)[/tex] is a root, then its complex conjugate [tex]\( a - bi \)[/tex] is also a root.
4. Given Information:
We are given that the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has one complex zero. Let's denote this complex zero as [tex]\( z_1 = a + bi \)[/tex] (where [tex]\( i \)[/tex] is the imaginary unit).
5. Determining the Other Zero:
Since the coefficients [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] are real numbers, and we know that complex zeros must come in conjugate pairs, the other zero of the quadratic equation must be the complex conjugate of [tex]\( z_1 \)[/tex]. Therefore, the second zero is [tex]\( z_2 = a - bi \)[/tex].
6. Conclusion:
The quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] indeed has another zero because it must have two solutions in total. According to the rules about the nature of complex zeros, the equation will have exactly 2 zeros.
Therefore, the quadratic equation has 2 zeros.
### Step-by-Step Solution:
1. Degree of the Polynomial:
A quadratic equation is a polynomial of degree 2. This means that the highest power of the variable [tex]\( x \)[/tex] is 2.
2. Fundamental Theorem of Algebra:
The Fundamental Theorem of Algebra states that every non-zero polynomial equation of degree [tex]\( n \)[/tex] has exactly [tex]\( n \)[/tex] solutions in the set of complex numbers, counting multiplicities. Since the degree of our quadratic equation is 2, it must have exactly 2 solutions in the complex number set.
3. Nature of Complex Zeros:
Complex roots (or zeros) of polynomials with real coefficients always occur in conjugate pairs. This means if [tex]\( a + bi \)[/tex] is a root, then its complex conjugate [tex]\( a - bi \)[/tex] is also a root.
4. Given Information:
We are given that the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has one complex zero. Let's denote this complex zero as [tex]\( z_1 = a + bi \)[/tex] (where [tex]\( i \)[/tex] is the imaginary unit).
5. Determining the Other Zero:
Since the coefficients [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] are real numbers, and we know that complex zeros must come in conjugate pairs, the other zero of the quadratic equation must be the complex conjugate of [tex]\( z_1 \)[/tex]. Therefore, the second zero is [tex]\( z_2 = a - bi \)[/tex].
6. Conclusion:
The quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] indeed has another zero because it must have two solutions in total. According to the rules about the nature of complex zeros, the equation will have exactly 2 zeros.
Therefore, the quadratic equation has 2 zeros.
