Answer:
Learning Task 2:
- 3x² - 2x + 5 = 0; a = 3, b = -2, and c = 5
- x² + 5x - 4 = 0; a = 1, b = 5, and c = -4
- 2x² - 3x - 20 = 0; a = 2, b = -3, and c = -20
- 2x² - 2x - 6 = 0; a = 2, b = -2, and c = -6
- x² + 5x - 4 = 0; a = 1, b = 5, and c = -4
Solutions:
1. 2x - 3x² = 5
Equate everything to 0 by adding -5 to both sides of the equation.
2x - 3x² + (-5) = 5 + (-5)
2x - 3x² - 5 = 0
Rearrange the terms of the equation to its standard form.
-3x² + 2x - 5 = 0
Remove the negative sign of the first term by multiplying both sides of the equation to -1.
-1(-3x² + 2x -5 = 0)-1
3x² - 2x + 5 = 0
Find the values of a, b, and c.
3x² - 2x + 5 = 0
a = 3, b = -2, and c = 5
2. 4 - x² = 5x
Equate everything to 0 by adding -5x to both sides of the equation.
4 - x² + (-5x) = 5x + (-5x)
4 - x² - 5x = 0
Rearrange the terms of the equation into its standard form.
-x² - 5x + 4 = 0
Remove the negative sign of the first term by multiplying both sides of the equation to -1.
-1(-x² - 5x + 4 = 0)-1
x² + 5x - 4 = 0
Find the values of a, b, and c.
x² + 5x - 4 = 0
a = 1, b = 5, and c = -4
3. (2x + 5) (x - 4) = 6
Multiply (2x + 5)(x - 4).
(2x + 5)(x - 4) = (2x)(x) + (2x)(-4) + (5)(x) + (5)(-4) = 2x² - 8x + 5x - 20 = 2x² - 3x - 20
Then, 2x² - 3x - 20 = 0
Give the values of a, b, and c.
a = 2, b = -3, and c = -20
4. 2x(x - 1) = 6
Solve for 2x(x - 1).
2x(x - 1) = (2x)(x) + (2x)(-1) = 2x² - 2x
Then, 2x² - 2x = 6
Equate everything to 0 by adding -6 to both sides of the equation.
2x² - 2x + (-6) = 6 + (-6)
2x² - 2x - 6 = 0
Give the values of a, b, and c.
2x² - 2x - 6 = 0
a = 2, b = -2, and c = -6
5. (x + 1)(x + 4) = 8
Solve for the product of (x + 1)(x + 4).
(x + 1)(x + 4) = (x)(x) + (x)(4) + (1)(x) + (1)(4) = x² + 4x + x + 4 = x² + 5x + 4
Then, x² + 5x + 4 = 8
Equate everything to 0 by adding -8 to both sides of the equation.
x² + 5x + 4 + (-8) = 8 + (-8)
x² + 5x - 4 = 0
Give the values of a, b, and c.
x² + 5x - 4 = 0
a = 1, b = 5, and c = -4
Keywords: equation, standard form
Explanation:
