Sure, let's solve the quadratic equation using the given values: [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -3\)[/tex].
The quadratic formula for finding the roots of the equation [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
To find the first solution, we will use only the positive square root part ([tex]\(+\sqrt{b^2 - 4ac}\)[/tex]) of the formula. Let's proceed step-by-step:
1. Calculate the Discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac
\][/tex]
Substituting the values [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -3\)[/tex]:
[tex]\[
\text{Discriminant} = 2^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16
\][/tex]
2. Calculate the square root of the Discriminant:
[tex]\[
\sqrt{\text{Discriminant}} = \sqrt{16} = 4
\][/tex]
3. Calculate the Numerator:
[tex]\[
\text{Numerator} = -b + \sqrt{\text{Discriminant}}
\][/tex]
Substituting [tex]\(b = 2\)[/tex] and [tex]\(\sqrt{\text{Discriminant}} = 4\)[/tex]:
[tex]\[
\text{Numerator} = -2 + 4 = 2
\][/tex]
4. Calculate the Denominator:
[tex]\[
\text{Denominator} = 2a
\][/tex]
Substituting [tex]\(a = 1\)[/tex]:
[tex]\[
\text{Denominator} = 2 \cdot 1 = 2
\][/tex]
5. Calculate the First Solution:
[tex]\[
x = \frac{\text{Numerator}}{\text{Denominator}}
\][/tex]
Substituting the Numerator [tex]\(= 2\)[/tex] and the Denominator [tex]\(= 2\)[/tex]:
[tex]\[
x = \frac{2}{2} = 1
\][/tex]
Therefore, the first solution to the quadratic equation given [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -3\)[/tex] is:
[tex]\[
x = 1
\][/tex]
Summarizing, we have:
- The discriminant is [tex]\(16\)[/tex].
- The numerator (without the denominator) is [tex]\(2\)[/tex].
- The first solution to the quadratic equation is [tex]\(1\)[/tex].
