Certainly! Let's go through each quadratic equation step-by-step to convert them into standard form [tex]\(ax^2 + bx + c = 0\)[/tex].
### Equation 1: [tex]\(3x - 2x^2 = 7\)[/tex]
1. Start by moving all terms to one side of the equation:
[tex]\[-2x^2 + 3x - 7 = 0\][/tex]
2. Here, [tex]\(a = -2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -7\)[/tex].
### Equation 2: [tex]\(5 - 2x^2 = 6x\)[/tex]
1. Move all terms to one side to form a standard quadratic equation:
[tex]\[
-2x^2 - 6x + 5 = 0
\][/tex]
2. Here, [tex]\(a = -2\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = 5\)[/tex].
### Equation 3: [tex]\((x + 3)(x + 4) = 0\)[/tex]
1. First, expand the left-hand side:
[tex]\[
x^2 + 4x + 3x + 12 = 0 \implies x^2 + 7x + 12 = 0
\][/tex]
2. Here, [tex]\(a = 1\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = 12\)[/tex].
### Equation 4: [tex]\((2x + 7)(x - 1) = 0\)[/tex]
1. Expand the left-hand side:
[tex]\[
2x^2 - 2x + 7x - 7 = 0 \implies 2x^2 + 5x - 7 = 0
\][/tex]
2. Here, [tex]\(a = 2\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -7\)[/tex].
### Equation 5: [tex]\(2x(x - 3) = 15\)[/tex]
1. First, distribute the [tex]\(2x\)[/tex]:
[tex]\[
2x^2 - 6x = 15
\][/tex]
2. Move all terms to one side to obtain:
[tex]\[
2x^2 - 6x - 15 = 0
\][/tex]
3. Here, [tex]\(a = 2\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = -15\)[/tex].
### Summary of Results
For each equation, the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are as follows:
1. [tex]\(3x - 2x^2 = 7\)[/tex] => [tex]\(a = -2\)[/tex], [tex]\(b = 3\)[/tex], [tex]\(c = -7\)[/tex]
2. [tex]\(5 - 2x^2 = 6x\)[/tex] => [tex]\(a = -2\)[/tex], [tex]\(b = -6\)[/tex], [tex]\(c = 5\)[/tex]
3. [tex]\((x + 3)(x + 4) = 0\)[/tex] => [tex]\(a = 1\)[/tex], [tex]\(b = 7\)[/tex], [tex]\(c = 12\)[/tex]
4. [tex]\((2x + 7)(x - 1) = 0\)[/tex] => [tex]\(a = 2\)[/tex], [tex]\(b = 5\)[/tex], [tex]\(c = -7\)[/tex]
5. [tex]\(2x(x - 3) = 15\)[/tex] => [tex]\(a = 2\)[/tex], [tex]\(b = -6\)[/tex], [tex]\(c = -15\)[/tex]
These are the standard forms for each quadratic equation.
