IDNLearn.com is designed to help you find the answers you need quickly and easily. Our community is here to provide the comprehensive and accurate answers you need to make informed decisions.
Sagot :
Sure, let's solve each quadratic equation step by step.
### 1. Solve the equation [tex]\(2 f^2 - 10 f + 8 = 0\)[/tex]:
To solve the quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex]:
1. Identify the coefficients: [tex]\(a = 2\)[/tex], [tex]\(b = -10\)[/tex], and [tex]\(c = 8\)[/tex].
2. Use the quadratic formula: [tex]\(f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
[tex]\[ b^2 - 4ac = (-10)^2 - 4 \cdot 2 \cdot 8 = 100 - 64 = 36 \][/tex]
[tex]\[ f = \frac{10 \pm \sqrt{36}}{4} = \frac{10 \pm 6}{4} \][/tex]
So,
[tex]\[ f_1 = \frac{10 + 6}{4} = 4 \][/tex]
[tex]\[ f_2 = \frac{10 - 6}{4} = 1 \][/tex]
The solutions are [tex]\(f = 1\)[/tex] and [tex]\(f = 4\)[/tex].
### 2. Solve the equation [tex]\(3 x^2 - 5 x + 4 = 0\)[/tex]:
1. Rearrange terms if necessary and identify the coefficients: [tex]\(a = 3\)[/tex], [tex]\(b = -5\)[/tex], [tex]\(c = 4\)[/tex].
2. Use the quadratic formula: [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
[tex]\[ b^2 - 4ac = (-5)^2 - 4 \cdot 3 \cdot 4 = 25 - 48 = -23 \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{-23}}{6} = \frac{5 \pm \sqrt{23}i}{6} \][/tex]
The solutions are:
[tex]\[ x_1 = \frac{5 - \sqrt{23}i}{6} \][/tex]
[tex]\[ x_2 = \frac{5 + \sqrt{23}i}{6} \][/tex]
### 3. Solve the equation [tex]\(r^2 + 6 r + 9 = 0\)[/tex]:
1. Identify the coefficients: [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = 9\)[/tex].
2. Use the quadratic formula: [tex]\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
Since this is a perfect square trinomial, we can also see,
[tex]\[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \][/tex]
[tex]\[ r = \frac{-6 \pm \sqrt{0}}{2} = \frac{-6}{2} = -3 \][/tex]
The solution is [tex]\(r = -3\)[/tex]. Note that this is a repeated root.
### 4. Solve the equation [tex]\(l^2 + 5 l + 10 = 0\)[/tex]:
1. Identify the coefficients: [tex]\(a = 1\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 10\)[/tex].
2. Use the quadratic formula: [tex]\(l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
[tex]\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 10 = 25 - 40 = -15 \][/tex]
[tex]\[ l = \frac{-5 \pm \sqrt{-15}}{2} = \frac{-5 \pm \sqrt{15}i}{2} \][/tex]
The solutions are:
[tex]\[ l_1 = \frac{-5 - \sqrt{15}i}{2} \][/tex]
[tex]\[ l_2 = \frac{-5 + \sqrt{15}i}{2} \][/tex]
So, to summarize, the solutions for each quadratic equation are:
1. [tex]\( f = 1 \)[/tex], [tex]\( f = 4 \)[/tex]
2. [tex]\( x = \frac{5 - \sqrt{23}i}{6} \)[/tex], [tex]\( x = \frac{5 + \sqrt{23}i}{6} \)[/tex]
3. [tex]\( r = -3 \)[/tex]
4. [tex]\( l = \frac{-5 - \sqrt{15}i}{2} \)[/tex], [tex]\( l = \frac{-5 + \sqrt{15}i}{2} \)[/tex]
### 1. Solve the equation [tex]\(2 f^2 - 10 f + 8 = 0\)[/tex]:
To solve the quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex]:
1. Identify the coefficients: [tex]\(a = 2\)[/tex], [tex]\(b = -10\)[/tex], and [tex]\(c = 8\)[/tex].
2. Use the quadratic formula: [tex]\(f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
[tex]\[ b^2 - 4ac = (-10)^2 - 4 \cdot 2 \cdot 8 = 100 - 64 = 36 \][/tex]
[tex]\[ f = \frac{10 \pm \sqrt{36}}{4} = \frac{10 \pm 6}{4} \][/tex]
So,
[tex]\[ f_1 = \frac{10 + 6}{4} = 4 \][/tex]
[tex]\[ f_2 = \frac{10 - 6}{4} = 1 \][/tex]
The solutions are [tex]\(f = 1\)[/tex] and [tex]\(f = 4\)[/tex].
### 2. Solve the equation [tex]\(3 x^2 - 5 x + 4 = 0\)[/tex]:
1. Rearrange terms if necessary and identify the coefficients: [tex]\(a = 3\)[/tex], [tex]\(b = -5\)[/tex], [tex]\(c = 4\)[/tex].
2. Use the quadratic formula: [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
[tex]\[ b^2 - 4ac = (-5)^2 - 4 \cdot 3 \cdot 4 = 25 - 48 = -23 \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{-23}}{6} = \frac{5 \pm \sqrt{23}i}{6} \][/tex]
The solutions are:
[tex]\[ x_1 = \frac{5 - \sqrt{23}i}{6} \][/tex]
[tex]\[ x_2 = \frac{5 + \sqrt{23}i}{6} \][/tex]
### 3. Solve the equation [tex]\(r^2 + 6 r + 9 = 0\)[/tex]:
1. Identify the coefficients: [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = 9\)[/tex].
2. Use the quadratic formula: [tex]\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
Since this is a perfect square trinomial, we can also see,
[tex]\[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \][/tex]
[tex]\[ r = \frac{-6 \pm \sqrt{0}}{2} = \frac{-6}{2} = -3 \][/tex]
The solution is [tex]\(r = -3\)[/tex]. Note that this is a repeated root.
### 4. Solve the equation [tex]\(l^2 + 5 l + 10 = 0\)[/tex]:
1. Identify the coefficients: [tex]\(a = 1\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 10\)[/tex].
2. Use the quadratic formula: [tex]\(l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
[tex]\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 10 = 25 - 40 = -15 \][/tex]
[tex]\[ l = \frac{-5 \pm \sqrt{-15}}{2} = \frac{-5 \pm \sqrt{15}i}{2} \][/tex]
The solutions are:
[tex]\[ l_1 = \frac{-5 - \sqrt{15}i}{2} \][/tex]
[tex]\[ l_2 = \frac{-5 + \sqrt{15}i}{2} \][/tex]
So, to summarize, the solutions for each quadratic equation are:
1. [tex]\( f = 1 \)[/tex], [tex]\( f = 4 \)[/tex]
2. [tex]\( x = \frac{5 - \sqrt{23}i}{6} \)[/tex], [tex]\( x = \frac{5 + \sqrt{23}i}{6} \)[/tex]
3. [tex]\( r = -3 \)[/tex]
4. [tex]\( l = \frac{-5 - \sqrt{15}i}{2} \)[/tex], [tex]\( l = \frac{-5 + \sqrt{15}i}{2} \)[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Discover the answers you need at IDNLearn.com. Thanks for visiting, and come back soon for more valuable insights.
