At IDNLearn.com, find answers to your most pressing questions from experts and enthusiasts alike. Get step-by-step guidance for all your technical questions from our knowledgeable community members.
Sagot :
To solve the quadratic equation [tex]\( x^2 - 5x + 2 = 0 \)[/tex], we will use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this equation, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]. For the given quadratic equation [tex]\(x^2 - 5x + 2 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -5\)[/tex]
- [tex]\(c = 2\)[/tex]
First, we need to calculate the discriminant, which is the part under the square root in the quadratic formula:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \text{Discriminant} = (-5)^2 - 4 \cdot 1 \cdot 2 \][/tex]
[tex]\[ \text{Discriminant} = 25 - 8 \][/tex]
[tex]\[ \text{Discriminant} = 17 \][/tex]
Since the discriminant is positive ([tex]\(17\)[/tex]), we will have two real roots. Now we substitute the coefficients and the discriminant into the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{17}}{2a} \][/tex]
Given [tex]\(a = 1\)[/tex] and [tex]\(b = -5\)[/tex]:
[tex]\[ x = \frac{-(-5) \pm \sqrt{17}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{17}}{2} \][/tex]
Now, we calculate the two roots separately:
1. For the plus sign ([tex]\(+\)[/tex]):
[tex]\[ x_1 = \frac{5 + \sqrt{17}}{2} \][/tex]
The numerical solution is approximately:
[tex]\[ x_1 \approx 4.561552812808831 \][/tex]
2. For the minus sign ([tex]\(-\)[/tex]):
[tex]\[ x_2 = \frac{5 - \sqrt{17}}{2} \][/tex]
The numerical solution is approximately:
[tex]\[ x_2 \approx 0.4384471871911697 \][/tex]
Thus, the roots of the quadratic equation [tex]\( x^2 - 5x + 2 = 0 \)[/tex] are approximately:
[tex]\[ x_1 \approx 4.561552812808831 \][/tex]
[tex]\[ x_2 \approx 0.4384471871911697 \][/tex]
These are the two solutions for the quadratic equation.
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this equation, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]. For the given quadratic equation [tex]\(x^2 - 5x + 2 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -5\)[/tex]
- [tex]\(c = 2\)[/tex]
First, we need to calculate the discriminant, which is the part under the square root in the quadratic formula:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \text{Discriminant} = (-5)^2 - 4 \cdot 1 \cdot 2 \][/tex]
[tex]\[ \text{Discriminant} = 25 - 8 \][/tex]
[tex]\[ \text{Discriminant} = 17 \][/tex]
Since the discriminant is positive ([tex]\(17\)[/tex]), we will have two real roots. Now we substitute the coefficients and the discriminant into the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{17}}{2a} \][/tex]
Given [tex]\(a = 1\)[/tex] and [tex]\(b = -5\)[/tex]:
[tex]\[ x = \frac{-(-5) \pm \sqrt{17}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{17}}{2} \][/tex]
Now, we calculate the two roots separately:
1. For the plus sign ([tex]\(+\)[/tex]):
[tex]\[ x_1 = \frac{5 + \sqrt{17}}{2} \][/tex]
The numerical solution is approximately:
[tex]\[ x_1 \approx 4.561552812808831 \][/tex]
2. For the minus sign ([tex]\(-\)[/tex]):
[tex]\[ x_2 = \frac{5 - \sqrt{17}}{2} \][/tex]
The numerical solution is approximately:
[tex]\[ x_2 \approx 0.4384471871911697 \][/tex]
Thus, the roots of the quadratic equation [tex]\( x^2 - 5x + 2 = 0 \)[/tex] are approximately:
[tex]\[ x_1 \approx 4.561552812808831 \][/tex]
[tex]\[ x_2 \approx 0.4384471871911697 \][/tex]
These are the two solutions for the quadratic equation.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.