f) 5x^2 - 3x - 2 = 0
g) 3x^2 + 55 = 0
h) x^2 - 6x = 0
i) x^2 - 10x + 25 = 0

Q: f) 5x^2 - 3x - 2 = 0 g) 3x^2 + 55 = 0 h) x^2 - 6x = 0 i) x^2 - 10x + 25 = 0

Let's solve each quadratic equation: f) $5x^2 - 3x - 2 = 0$ To solve, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $ a = 5 $, $ b = -3 $, and $ c = -2 $. \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5} \] \[ x = \frac{3 \pm \sqrt{9 + 40}}{10} \] \[ x = \frac{3 \pm \sqrt{49}}{10} \] \[ x = \frac{3 \pm 7}{10} \] Solutions: \[ x = 1 \quad \text{or} \quad x = -\frac{2}{5} \] g) $3x^2 + 55 = 0$ This equation has no real solutions since the discriminant is negative. \[ 3x^2 = -55 \] h) $x^2 - 6x = 0$ Factor the equation: \[ x(x - 6) = 0 \] Solutions: \[ x = 0 \quad \text{or} \quad x = 6 \] i) $x^2 - 10x + 25 = 0$ This can be factored as: \[ (x - 5)^2 = 0 \] Solution: \[ x = 5 \]

Question

Accepted Answer

Let's solve each quadratic equation:

f) $5x^2 - 3x - 2 = 0$

To solve, use the quadratic formula: 
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
where $ a = 5 $, $ b = -3 $, and $ c = -2 $.

$$ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5} $$
$$ x = \frac{3 \pm \sqrt{9 + 40}}{10} $$
$$ x = \frac{3 \pm \sqrt{49}}{10} $$
$$ x = \frac{3 \pm 7}{10} $$

Solutions:
$$ x = 1 \quad \text{or} \quad x = -\frac{2}{5} $$

g) $3x^2 + 55 = 0$

This equation has no real solutions since the discriminant is negative.
$$ 3x^2 = -55 $$

h) $x^2 - 6x = 0$

Factor the equation:
$$ x(x - 6) = 0 $$

Solutions:
$$ x = 0 \quad \text{or} \quad x = 6 $$

i) $x^2 - 10x + 25 = 0$

This can be factored as:
$$ (x - 5)^2 = 0 $$

Solution:
$$ x = 5 $$

f) 5x^2 - 3x - 2 = 0 g) 3x^2 + 55 = 0 h) x^2 - 6x = 0 i) x^2...

f) 5x^2 - 3x - 2 = 0 g) 3x^2 + 55 = 0 h) x^2 - 6x = 0 i) x^2 - 10x + 25 = 0

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