Question

1. the solution of the equation \\(5x^2 - 2x - 39 = 0\\)

\\(3, 13/5\\)
\\(3, -13/5\\)
\\(3, -13\\)
\\(-3, 13/5\\)
\\(-3, -13/5\\)

Explanation:

Identify coefficients of the quadratic equation

$$ a = 5, \quad b = -2, \quad c = -39 $$

Apply the quadratic formula

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

$$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(-39)}}{2(5)} $$

$$ x = \frac{2 \pm \sqrt{4 + 780}}{10} $$

$$ x = \frac{2 \pm \sqrt{784}}{10} $$

Simplify the roots

$$ \sqrt{784} = 28 $$

$$ x_1 = \frac{2 + 28}{10} = 3 $$

$$ x_2 = \frac{2 - 28}{10} = -\frac{26}{10} = -\frac{13}{5} $$

Answer:

• 3, 13/5
• 3, -13/5 (Correct answer)
• 3, -13
• -3, 13/5
• -3, -13/5