Uncover valuable information and solutions with IDNLearn.com's extensive Q&A platform. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
To solve the quadratic equation [tex]\( 3x^2 + x - 5 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
where [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 equation [tex]\( 3x^2 + x - 5 = 0 \)[/tex], the coefficients are:
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = -5 \)[/tex]
Step-by-step solution:
1. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \text{Discriminant} = 1^2 - 4 \cdot 3 \cdot (-5) \][/tex]
[tex]\[ \text{Discriminant} = 1 - (-60) \][/tex]
[tex]\[ \text{Discriminant} = 1 + 60 \][/tex]
[tex]\[ \text{Discriminant} = 61 \][/tex]
2. Find the two solutions using the quadratic formula:
[tex]\[ x_1 = \frac{{-b + \sqrt{{\text{Discriminant}}}}}{2a} \][/tex]
[tex]\[ x_2 = \frac{{-b - \sqrt{{\text{Discriminant}}}}}{2a} \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and the discriminant:
[tex]\[ x_1 = \frac{{-1 + \sqrt{61}}}{2 \cdot 3} \][/tex]
[tex]\[ x_2 = \frac{{-1 - \sqrt{61}}}{2 \cdot 3} \][/tex]
3. Calculate the exact values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ x_1 = \frac{{-1 + \sqrt{61}}}{6} \][/tex]
[tex]\[ x_2 = \frac{{-1 - \sqrt{61}}}{6} \][/tex]
Numerically, these values are approximately:
[tex]\[ x_1 \approx 1.135041612651109 \][/tex]
[tex]\[ x_2 \approx -1.4683749459844424 \][/tex]
4. Round the solutions to 2 decimal places:
[tex]\[ x_1 \approx 1.14 \][/tex]
[tex]\[ x_2 \approx -1.47 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\( 3x^2 + x - 5 = 0 \)[/tex] rounded to two decimal places are [tex]\( x_1 = 1.14 \)[/tex] and [tex]\( x_2 = -1.47 \)[/tex].
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
where [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 equation [tex]\( 3x^2 + x - 5 = 0 \)[/tex], the coefficients are:
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = -5 \)[/tex]
Step-by-step solution:
1. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \text{Discriminant} = 1^2 - 4 \cdot 3 \cdot (-5) \][/tex]
[tex]\[ \text{Discriminant} = 1 - (-60) \][/tex]
[tex]\[ \text{Discriminant} = 1 + 60 \][/tex]
[tex]\[ \text{Discriminant} = 61 \][/tex]
2. Find the two solutions using the quadratic formula:
[tex]\[ x_1 = \frac{{-b + \sqrt{{\text{Discriminant}}}}}{2a} \][/tex]
[tex]\[ x_2 = \frac{{-b - \sqrt{{\text{Discriminant}}}}}{2a} \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and the discriminant:
[tex]\[ x_1 = \frac{{-1 + \sqrt{61}}}{2 \cdot 3} \][/tex]
[tex]\[ x_2 = \frac{{-1 - \sqrt{61}}}{2 \cdot 3} \][/tex]
3. Calculate the exact values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ x_1 = \frac{{-1 + \sqrt{61}}}{6} \][/tex]
[tex]\[ x_2 = \frac{{-1 - \sqrt{61}}}{6} \][/tex]
Numerically, these values are approximately:
[tex]\[ x_1 \approx 1.135041612651109 \][/tex]
[tex]\[ x_2 \approx -1.4683749459844424 \][/tex]
4. Round the solutions to 2 decimal places:
[tex]\[ x_1 \approx 1.14 \][/tex]
[tex]\[ x_2 \approx -1.47 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\( 3x^2 + x - 5 = 0 \)[/tex] rounded to two decimal places are [tex]\( x_1 = 1.14 \)[/tex] and [tex]\( x_2 = -1.47 \)[/tex].
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.