Answer :
Certainly! Let's solve each of the given quadratic equations step-by-step using the quadratic formula:
### Quadratic Formula
Given a quadratic equation in the form:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
The solutions for [tex]\(x\)[/tex] are given by the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### 1) [tex]\(m^2 - 5m - 14 = 0\)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = -14\)[/tex].
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 + 56}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 9}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{5 + 9}{2} = 7 \][/tex]
[tex]\[ x = \frac{5 - 9}{2} = -2 \][/tex]
So, the roots are [tex]\(7\)[/tex] and [tex]\(-2\)[/tex].
### 2) [tex]\(b^2 - 4b + 4 = 0\)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 4\)[/tex].
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 - 16}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{0}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 0}{2} \][/tex]
Thus, the only solution is:
[tex]\[ x = 2 \][/tex]
So, the root is [tex]\(2\)[/tex] (a repeated root).
### 3) [tex]\(2m^2 + 2m - 12 = 0\)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -12\)[/tex].
[tex]\[ x = \frac{-(2) \pm \sqrt{(2)^2 - 4 \cdot 2 \cdot (-12)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 96}}{4} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{100}}{4} \][/tex]
[tex]\[ x = \frac{-2 \pm 10}{4} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-2 + 10}{4} = 2 \][/tex]
[tex]\[ x = \frac{-2 - 10}{4} = -3 \][/tex]
So, the roots are [tex]\(2\)[/tex] and [tex]\(-3\)[/tex].
### 4) [tex]\(2x^2 - 3x - 5 = 0\)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -5\)[/tex].
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 40}}{4} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{49}}{4} \][/tex]
[tex]\[ x = \frac{3 \pm 7}{4} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{3 + 7}{4} = 2.5 \][/tex]
[tex]\[ x = \frac{3 - 7}{4} = -1 \][/tex]
So, the roots are [tex]\(2.5\)[/tex] and [tex]\(-1\)[/tex].
### 5) [tex]\(x^2 + 4x + 3 = 0\)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 3\)[/tex].
[tex]\[ x = \frac{-(4) \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-4 + 2}{2} = -1 \][/tex]
[tex]\[ x = \frac{-4 - 2}{2} = -3 \][/tex]
So, the roots are [tex]\(-1\)[/tex] and [tex]\(-3\)[/tex].
### 6) [tex]\(2x^2 + 3x - 20 = 0\)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -20\)[/tex].
[tex]\[ x = \frac{-(3) \pm \sqrt{(3)^2 - 4 \cdot 2 \cdot (-20)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 160}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{169}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm 13}{4} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-3 + 13}{4} = 2.5 \][/tex]
[tex]\[ x = \frac{-3 - 13}{4} = -4 \][/tex]
So, the roots are [tex]\(2.5\)[/tex] and [tex]\(-4\)[/tex].
### Summary of Solutions
1. [tex]\(m^2 - 5m - 14 = 0\)[/tex] → Roots: [tex]\(7\)[/tex] and [tex]\(-2\)[/tex]
2. [tex]\(b^2 - 4b + 4 = 0\)[/tex] → Roots: [tex]\(2\)[/tex] (repeated root)
3. [tex]\(2m^2 + 2m - 12 = 0\)[/tex] → Roots: [tex]\(2\)[/tex] and [tex]\(-3\)[/tex]
4. [tex]\(2x^2 - 3x - 5 = 0\)[/tex] → Roots: [tex]\(2.5\)[/tex] and [tex]\(-1\)[/tex]
5. [tex]\(x^2 + 4x + 3 = 0\)[/tex] → Roots: [tex]\(-1\)[/tex] and [tex]\(-3\)[/tex]
6. [tex]\(2x^2 + 3x - 20 = 0\)[/tex] → Roots: [tex]\(2.5\)[/tex] and [tex]\(-4\)[/tex]
These steps show how each quadratic equation is solved using the quadratic formula, resulting in the solutions provided.
### Quadratic Formula
Given a quadratic equation in the form:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
The solutions for [tex]\(x\)[/tex] are given by the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### 1) [tex]\(m^2 - 5m - 14 = 0\)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = -14\)[/tex].
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 + 56}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 9}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{5 + 9}{2} = 7 \][/tex]
[tex]\[ x = \frac{5 - 9}{2} = -2 \][/tex]
So, the roots are [tex]\(7\)[/tex] and [tex]\(-2\)[/tex].
### 2) [tex]\(b^2 - 4b + 4 = 0\)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 4\)[/tex].
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 - 16}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{0}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 0}{2} \][/tex]
Thus, the only solution is:
[tex]\[ x = 2 \][/tex]
So, the root is [tex]\(2\)[/tex] (a repeated root).
### 3) [tex]\(2m^2 + 2m - 12 = 0\)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -12\)[/tex].
[tex]\[ x = \frac{-(2) \pm \sqrt{(2)^2 - 4 \cdot 2 \cdot (-12)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 96}}{4} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{100}}{4} \][/tex]
[tex]\[ x = \frac{-2 \pm 10}{4} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-2 + 10}{4} = 2 \][/tex]
[tex]\[ x = \frac{-2 - 10}{4} = -3 \][/tex]
So, the roots are [tex]\(2\)[/tex] and [tex]\(-3\)[/tex].
### 4) [tex]\(2x^2 - 3x - 5 = 0\)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -5\)[/tex].
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 40}}{4} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{49}}{4} \][/tex]
[tex]\[ x = \frac{3 \pm 7}{4} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{3 + 7}{4} = 2.5 \][/tex]
[tex]\[ x = \frac{3 - 7}{4} = -1 \][/tex]
So, the roots are [tex]\(2.5\)[/tex] and [tex]\(-1\)[/tex].
### 5) [tex]\(x^2 + 4x + 3 = 0\)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 3\)[/tex].
[tex]\[ x = \frac{-(4) \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-4 + 2}{2} = -1 \][/tex]
[tex]\[ x = \frac{-4 - 2}{2} = -3 \][/tex]
So, the roots are [tex]\(-1\)[/tex] and [tex]\(-3\)[/tex].
### 6) [tex]\(2x^2 + 3x - 20 = 0\)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -20\)[/tex].
[tex]\[ x = \frac{-(3) \pm \sqrt{(3)^2 - 4 \cdot 2 \cdot (-20)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 160}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{169}}{4} \][/tex]
[tex]\[ x = \frac{-3 \pm 13}{4} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-3 + 13}{4} = 2.5 \][/tex]
[tex]\[ x = \frac{-3 - 13}{4} = -4 \][/tex]
So, the roots are [tex]\(2.5\)[/tex] and [tex]\(-4\)[/tex].
### Summary of Solutions
1. [tex]\(m^2 - 5m - 14 = 0\)[/tex] → Roots: [tex]\(7\)[/tex] and [tex]\(-2\)[/tex]
2. [tex]\(b^2 - 4b + 4 = 0\)[/tex] → Roots: [tex]\(2\)[/tex] (repeated root)
3. [tex]\(2m^2 + 2m - 12 = 0\)[/tex] → Roots: [tex]\(2\)[/tex] and [tex]\(-3\)[/tex]
4. [tex]\(2x^2 - 3x - 5 = 0\)[/tex] → Roots: [tex]\(2.5\)[/tex] and [tex]\(-1\)[/tex]
5. [tex]\(x^2 + 4x + 3 = 0\)[/tex] → Roots: [tex]\(-1\)[/tex] and [tex]\(-3\)[/tex]
6. [tex]\(2x^2 + 3x - 20 = 0\)[/tex] → Roots: [tex]\(2.5\)[/tex] and [tex]\(-4\)[/tex]
These steps show how each quadratic equation is solved using the quadratic formula, resulting in the solutions provided.
