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]
