Answer :
Certainly! We can solve these quadratic equations using the quadratic formula, which is given by:
[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 from the general form of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
Let's go through each of the equations step-by-step.
### 1. Solve [tex]\(5h^2 - 2h - 7 = 0\)[/tex]
The equation is already in the standard form [tex]\(ah^2 + bh + c = 0\)[/tex] with:
- [tex]\(a = 5\)[/tex]
- [tex]\(b = -2\)[/tex]
- [tex]\(c = -7\)[/tex]
Using the quadratic formula:
[tex]\[ h = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-7)}}{2 \cdot 5} \][/tex]
[tex]\[ h = \frac{2 \pm \sqrt{4 + 140}}{10} \][/tex]
[tex]\[ h = \frac{2 \pm \sqrt{144}}{10} \][/tex]
[tex]\[ h = \frac{2 \pm 12}{10} \][/tex]
Finding the two solutions:
[tex]\[ h_1 = \frac{2 + 12}{10} = \frac{14}{10} = 1.4 \][/tex]
[tex]\[ h_2 = \frac{2 - 12}{10} = \frac{-10}{10} = -1 \][/tex]
So, the solutions are [tex]\(h = 1.4\)[/tex] and [tex]\(h = -1\)[/tex].
### 2. Solve [tex]\(6x^2 + x - 15 = 0\)[/tex]
The equation is already in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with:
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = -15\)[/tex]
Using the quadratic formula:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-15)}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 360}}{12} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{361}}{12} \][/tex]
[tex]\[ x = \frac{-1 \pm 19}{12} \][/tex]
Finding the two solutions:
[tex]\[ x_1 = \frac{-1 + 19}{12} = \frac{18}{12} = 1.5 \][/tex]
[tex]\[ x_2 = \frac{-1 - 19}{12} = \frac{-20}{12} = -\frac{5}{3} \][/tex]
So, the solutions are [tex]\(x = 1.5\)[/tex] and [tex]\(x = -\frac{5}{3}\)[/tex].
### 3. Solve [tex]\(3x^2 + 25x - 18 = 0\)[/tex]
The equation is already in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 25\)[/tex]
- [tex]\(c = -18\)[/tex]
Using the quadratic formula:
[tex]\[ x = \frac{-25 \pm \sqrt{25^2 - 4 \cdot 3 \cdot (-18)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{625 + 216}}{6} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{841}}{6} \][/tex]
[tex]\[ x = \frac{-25 \pm 29}{6} \][/tex]
Finding the two solutions:
[tex]\[ x_1 = \frac{-25 + 29}{6} = \frac{4}{6} = \frac{2}{3} \][/tex]
[tex]\[ x_2 = \frac{-25 - 29}{6} = \frac{-54}{6} = -9 \][/tex]
So, the solutions are [tex]\(x = \frac{2}{3}\)[/tex] and [tex]\(x = -9\)[/tex].
### 4. Solve [tex]\(2x^2 - 7x + 6 = 0\)[/tex]
The equation is already in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = -7\)[/tex]
- [tex]\(c = 6\)[/tex]
Using the quadratic formula:
[tex]\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 - 48}}{4} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{1}}{4} \][/tex]
[tex]\[ x = \frac{7 \pm 1}{4} \][/tex]
Finding the two solutions:
[tex]\[ x_1 = \frac{7 + 1}{4} = \frac{8}{4} = 2 \][/tex]
[tex]\[ x_2 = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2} \][/tex]
So, the solutions are [tex]\(x = 2\)[/tex] and [tex]\(x = \frac{3}{2}\)[/tex].
In summary, the truth sets for the given equations are:
1. [tex]\(5h^2 - 2h - 7 = 0\)[/tex] yields [tex]\(h = -1, 1.4\)[/tex]
2. [tex]\(6x^2 + x - 15 = 0\)[/tex] yields [tex]\(x = -\frac{5}{3}, 1.5\)[/tex]
3. [tex]\(3x^2 + 25x - 18 = 0\)[/tex] yields [tex]\(x = -9, \frac{2}{3}\)[/tex]
4. [tex]\(2x^2 - 7x + 6 = 0\)[/tex] yields [tex]\(x = 2, \frac{3}{2}\)[/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 from the general form of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
Let's go through each of the equations step-by-step.
### 1. Solve [tex]\(5h^2 - 2h - 7 = 0\)[/tex]
The equation is already in the standard form [tex]\(ah^2 + bh + c = 0\)[/tex] with:
- [tex]\(a = 5\)[/tex]
- [tex]\(b = -2\)[/tex]
- [tex]\(c = -7\)[/tex]
Using the quadratic formula:
[tex]\[ h = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-7)}}{2 \cdot 5} \][/tex]
[tex]\[ h = \frac{2 \pm \sqrt{4 + 140}}{10} \][/tex]
[tex]\[ h = \frac{2 \pm \sqrt{144}}{10} \][/tex]
[tex]\[ h = \frac{2 \pm 12}{10} \][/tex]
Finding the two solutions:
[tex]\[ h_1 = \frac{2 + 12}{10} = \frac{14}{10} = 1.4 \][/tex]
[tex]\[ h_2 = \frac{2 - 12}{10} = \frac{-10}{10} = -1 \][/tex]
So, the solutions are [tex]\(h = 1.4\)[/tex] and [tex]\(h = -1\)[/tex].
### 2. Solve [tex]\(6x^2 + x - 15 = 0\)[/tex]
The equation is already in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with:
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = -15\)[/tex]
Using the quadratic formula:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-15)}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 360}}{12} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{361}}{12} \][/tex]
[tex]\[ x = \frac{-1 \pm 19}{12} \][/tex]
Finding the two solutions:
[tex]\[ x_1 = \frac{-1 + 19}{12} = \frac{18}{12} = 1.5 \][/tex]
[tex]\[ x_2 = \frac{-1 - 19}{12} = \frac{-20}{12} = -\frac{5}{3} \][/tex]
So, the solutions are [tex]\(x = 1.5\)[/tex] and [tex]\(x = -\frac{5}{3}\)[/tex].
### 3. Solve [tex]\(3x^2 + 25x - 18 = 0\)[/tex]
The equation is already in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 25\)[/tex]
- [tex]\(c = -18\)[/tex]
Using the quadratic formula:
[tex]\[ x = \frac{-25 \pm \sqrt{25^2 - 4 \cdot 3 \cdot (-18)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{625 + 216}}{6} \][/tex]
[tex]\[ x = \frac{-25 \pm \sqrt{841}}{6} \][/tex]
[tex]\[ x = \frac{-25 \pm 29}{6} \][/tex]
Finding the two solutions:
[tex]\[ x_1 = \frac{-25 + 29}{6} = \frac{4}{6} = \frac{2}{3} \][/tex]
[tex]\[ x_2 = \frac{-25 - 29}{6} = \frac{-54}{6} = -9 \][/tex]
So, the solutions are [tex]\(x = \frac{2}{3}\)[/tex] and [tex]\(x = -9\)[/tex].
### 4. Solve [tex]\(2x^2 - 7x + 6 = 0\)[/tex]
The equation is already in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = -7\)[/tex]
- [tex]\(c = 6\)[/tex]
Using the quadratic formula:
[tex]\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 - 48}}{4} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{1}}{4} \][/tex]
[tex]\[ x = \frac{7 \pm 1}{4} \][/tex]
Finding the two solutions:
[tex]\[ x_1 = \frac{7 + 1}{4} = \frac{8}{4} = 2 \][/tex]
[tex]\[ x_2 = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2} \][/tex]
So, the solutions are [tex]\(x = 2\)[/tex] and [tex]\(x = \frac{3}{2}\)[/tex].
In summary, the truth sets for the given equations are:
1. [tex]\(5h^2 - 2h - 7 = 0\)[/tex] yields [tex]\(h = -1, 1.4\)[/tex]
2. [tex]\(6x^2 + x - 15 = 0\)[/tex] yields [tex]\(x = -\frac{5}{3}, 1.5\)[/tex]
3. [tex]\(3x^2 + 25x - 18 = 0\)[/tex] yields [tex]\(x = -9, \frac{2}{3}\)[/tex]
4. [tex]\(2x^2 - 7x + 6 = 0\)[/tex] yields [tex]\(x = 2, \frac{3}{2}\)[/tex]
