find the roots/zeros of the following quadratic equations using any method.
with solution
[tex]x {}^{2} - 6x + 5 = 0[/tex]
[tex]3x {}^{2} - 12x - 15 = 0[/tex]
[tex](2x + 3)(x - 4) = 0[/tex]
[tex]x {}^{2} - 12 - 85 = 0[/tex]
[tex]2x {}^{2} + 28x + 48 = 0[/tex]
[tex]x {}^{2} + 2x - 3 = 0 \\[/tex]
[tex](x - 5)(x + 3) = 0[/tex]
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Answer:
1. x = 5; x = 1
2. x = 5; x = - 1
3. x = ⅔; x = 4
4. x = 17; x = -5
5. x = 12; x = 2
6. x = - 3 ; x = 1
7. x = 5; x = - 3
Step-by-step explanation:
1. x² - 6x + 5 = 0
Use factoring to solve the equation.
(x - 5)(x - 1) = 0
Make two separate cases.
x - 5 = 0 x - 1 = 0
Transpose the constant value to the right side of the equation.
x = 5 ; x = 1
2. 3x² - 12x - 15 = 0
Use factoring to solve the equation.
(3x - 15)(x + 1) = 0
Make two separate cases.
3x - 15 = 0 x + 1 = 0
Transpose the constant value.
3x = 15 x = - 1
Divide 3x and 15 into 3.
x = 5 x = - 1
3. (2x + 3)(x - 4) = 0
Make two separate cases.
2x + 3 = 0 x - 4 = 0
Transpose the constant value.
2x = 3 x = 4
Divide 2x and 3 into 2.
x = ⅔ x = 4
4. x² - 12 - 85 (You mean x² - 12x - 85? I will proceed this one.)
Use factoring to solve the equation.
(x - 17)(x + 5) = 0
Make two separate cases.
x - 17 = 0 x + 5 = 0
Transpose the constant value.
x = 17 x = - 5
5. 2x² - 28x + 48 = 0
Factor the equation.
(2x - 24)(x - 2) = 0
Make two cases.
2x - 24 = 0 x - 2 = 0
Transpose as usual.
2x = 24 x = 2
Divide 2x and 24 into 2.
x = 12 x = 2
6. x² + 2x - 3 = 0
Factor the equation.
(x + 3)(x - 1) = 0
Make 2 cases.
x + 3 = 0 x - 1 = 0
Transpose the constant value.
x = - 3 x = 1
7. (x - 5)(x + 3) = 0
Make two cases.
x - 5 = 0 x + 3 = 0
Transpose as usual.
x = 5 x = - 3
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