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Andrew [12]
4 years ago
9

Both of these please and thank you!!!!!

Mathematics
1 answer:
Tom [10]4 years ago
7 0

Answer:

<u>1. 3x^2 + 4x - 5 = 0</u>

<u>x₁ = (-2 + √19)/3</u>

<u>x₂ = (-2 - √19)/3</u>

<u>2. X^2 - 13x + 42 = 0</u>

<u>x₁ = 7</u>

<u>x₂ = 6</u>

Step-by-step explanation:

1. Let's solve for x using the quadratic formula this way:

3x^2 + 4x - 5 = 0

x = [-4 +/- √(4² - 4 * 3 * -5)]/2 * 3

x = [-4 +/- √(16 + 60)]/6

x = [-4 +/- √76]/6

x = [-4 +/- 2√19]/6 (√76 =√4 * 19)

x = -2 +/- √19/3

<u>x₁ = (-2 + √19)/3</u>

<u>x₂ = (-2 - √19)/3</u>

2. Let's solve for x, factoring this way:

X^2 - 13x + 42 = 0

(x - 7) (x - 6) = 0

<u>x₁ = 7</u>

<u>x₂ = 6</u>

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Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might originally have been positive and might originally have been negative, I must acknowledge this fact when I remove the absolute-value bars. I do this by splitting the equation into two cases. For this exercise, these cases are as follows:

a. If the value of x was non-negative (that is, if it was positive or zero) to start with, then I can bring that value out of the absolute-value bars without changing its sign, giving me the equation x = 3.

b. If the value of x was negative to start with, then I can bring that value out of the absolute-value bars by changing the sign on x, giving me the equation –x = 3, which solves as x = –3.

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